Solutions of Triangles
At the end of this lecture, a student must be able to: 1 Use trigonometric functions to solve right triangles
2 Apply the laws of sines and cosines to solve oblique triangles
3 Use the trigonometric functions in solving real world triangles
Triangles
Notation for triangles
1 A, B, C: vertices
2 a, b, c: lengths of the respective opposite sides
Triangles
1 A, B, C: vertices
2 a, b, c: lengths of the respective opposite sides
3 α, β,γ: measures of the respective interior angles
Recall:
α+β+γ = 180◦
Triangle InequalityThe sum of the lengths of two sides of a triangle is greater than the length of the third side.
Hinge TheoremIn a 4, the side opposite the larger angle is the longer side.
Definition
A triangle that is not a right triangle is called anoblique triangle.
GOAL:
Definition
A triangle that is not a right triangle is called anoblique triangle.
GOAL:
Solving Right Triangles
θ: acute angle in a right triangle
Place the4on the Cartesian plane such thatθ is in standard position
θ
adjacent
opp
os
ite
hyp oten
use
x= adjacent,y= opposite, r= hypotenuse sinθ = opposite
hypotenuse, cosθ =
adjacent
hypotenuse, tanθ =
Solving Right Triangles
θ: acute angle in a right triangle
Place the4on the Cartesian plane such thatθ is in standard position
θ
adjacent
opp
os
ite
hyp oten
use
x= adjacent,y= opposite, r= hypotenuse sinθ = opposite
hypotenuse, cosθ =
adjacent
hypotenuse, tanθ =
Solving Right Triangles
θ: acute angle in a right triangle
Place the4on the Cartesian plane such thatθ is in standard position
θ
adjacent
opp
os
ite
hyp oten
use
x= adjacent,y= opposite, r= hypotenuse
sinθ = opposite
hypotenuse, cosθ =
adjacent
hypotenuse, tanθ =
Solving Right Triangles
θ: acute angle in a right triangle
Place the4on the Cartesian plane such thatθ is in standard position
θ
adjacent
opp
os
ite
hyp oten
use
x= adjacent,y= opposite, r= hypotenuse sinθ = opposite
hypotenuse
, cosθ = adjacent
hypotenuse, tanθ =
Solving Right Triangles
θ: acute angle in a right triangle
Place the4on the Cartesian plane such thatθ is in standard position
θ
adjacent
opp
os
ite
hyp oten
use
x= adjacent,y= opposite, r= hypotenuse sinθ = opposite
hypotenuse, cosθ =
adjacent hypotenuse
Solving Right Triangles
θ: acute angle in a right triangle
Place the4on the Cartesian plane such thatθ is in standard position
θ
adjacent
opp
os
ite
hyp oten
use
x= adjacent,y= opposite, r= hypotenuse sinθ = opposite
hypotenuse, cosθ =
adjacent
hypotenuse, tanθ =
Solving Right Triangles
Theorem
For an acute angle θ in a right triangle, we have
sinθ = opposite length hypotenuse length
cosθ = adjacent length hypotenuse length
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦ −(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution: α=
180◦ −(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦)
=75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ =
c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦
= 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦)
= 12 (sin 45◦cos 30◦−cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 ·1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ =
a 12
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦
= 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦)
= 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
=
Solving Right Triangles
Example: Solve the triangle withb = 12, γ = 15◦, and
β = 90◦.
Solution:
α= 180◦−(90◦+ 15◦) = 75◦
sin 15◦ = c 12
=⇒ c= 12 sin 15◦ = 12 sin(45◦−30◦) = 12 (sin 45◦cos 30◦ −cos 45◦sin 30◦)
= 12 √ 2 2 · √ 3 2 − √ 2 2 · 1 2 = 12 √ 6−√2
4
=3√6−3√2
cos 15◦ = 12a
=⇒ a= 12 cos 15◦ = 12 cos(45◦−30◦) = 12 (cos 45◦cos 30◦+ sin 45◦sin 30◦)
= 12
√ 6+√2
4
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB.
Then,sinα= hb and sinβ = ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα
= hb and sinβ = ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb
and sinβ = ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ
= ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha.
Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=
bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=bsinα
= asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=bsinα = asinβ
sinα a =
sinβ b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=bsinα = asinβ
sinα
a =
sinβ b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Sine Law
Drop a perpendicular segment from C to AB. Then,sinα= hb and sinβ = ha. Hence,
h=bsinα = asinβ sinα
a = sinβ
b
Theorem (Sine Law)
For a triangle with interior anglesα, β, γ opposite sides with lengths
a, b, crespectively: sinα
a =
sinβ b =
sinγ c
or a
sinα = b
sinβ = c
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b
⇒b= 8 sin 60
◦
sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=
12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a
⇒a= 8 sin 45
◦
sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
=
Example: Solve the triangle withc= 8, α= 45◦, and β = 60◦.
γ = 180◦−(60◦+ 45◦) = 75◦
sin 75◦
8 =
sin 60◦
b ⇒b=
8 sin 60◦ sin 75◦ =
8 √ 3 2 √ 6+√2
4
=12√2−4√6
sin 75◦
8 =
sin 45◦
a ⇒a=
8 sin 45◦ sin 75◦ =
8 √ 2 2 √ 6+√2
4
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2
= sin 60
◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦ 5√3
⇒ sinβ = 5
√ 2 √ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3
=
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√
2
2 and β ∈[0
◦,180◦).
Eitherβ = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦.
But if β = 135◦,β+γ >180◦. So, β =45◦ α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦.
So, β =45◦ α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦)
=75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3 sin 60◦
⇒a= 5
√
3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Solve all triangles ABC satisfyingb = 5√2, c= 5√3, γ = 60◦, if any.
sinβ
5√2 =
sin 60◦
5√3 ⇒ sinβ = 5√2
√ 3 2
5√3 =
√
2 2
sinβ =
√ 2
2 and β ∈[0
◦,180◦). Either β = 45◦ or β = 135◦. But if β = 135◦,β+γ >180◦. So, β =45◦
α= 180◦−(45◦+ 60◦) =75◦
a
sin 75◦ =
5√3
sin 60◦ ⇒a=
5√3
√ 6+√2
4 √ 3 2 = 5 √
Cosine Law
Set4ABC on the Cartesian plane as shown below:
Let(x, y) be the coordinates of A. Then
cosγ = xb and sinγ = yb
Cosine Law
Set4ABC on the Cartesian plane as shown below:
Let(x, y) be the coordinates ofA. Then
cosγ = xb and sinγ = yb
Cosine Law
Set4ABC on the Cartesian plane as shown below:
Let(x, y) be the coordinates ofA. Then
cosγ = xb and sinγ = yb
Cosine Law
Set4ABC on the Cartesian plane as shown below:
Let(x, y) be the coordinates ofA. Then
cosγ = xb and sinγ = yb
Cosine Law
Set4ABC on the Cartesian plane as shown below:
Let(x, y) be the coordinates ofA. Then
cosγ = xb and sinγ = yb
Cosine Law
By the distance formula,
c2 = (bcosγ−a)2+ (bsinγ−0)2
Cosine Law
By the distance formula,
c2 = (bcosγ−a)2+ (bsinγ−0)2
c2 = (b2cos2γ−2abcosγ+a2) +b2sin2γ
Cosine Law
By the distance formula,
c2 = (bcosγ−a)2+ (bsinγ−0)2
c2 = (b2cos2γ−2abcosγ+a2)+b2sin2γ
c2 = a2+b2(sin2γ+ cos2γ)−2abcosγ
Cosine Law
By the distance formula,
c2 = (bcosγ−a)2+ (bsinγ−0)2
Cosine Law
By changing the side of the triangle fixed on the positive x-axis, we get:
Theorem (Cosine Law)
a2 =b2+c2−2bccosα
b2 =a2+c2 −2accosβ
c2 =a2+b2−2abcosγ
Note:
Cosine Law
By changing the side of the triangle fixed on the positive x-axis, we get:
Theorem (Cosine Law)
a2 =b2+c2−2bccosα
b2 =a2+c2 −2accosβ
c2 =a2+b2−2abcosγ
Note:
When one of the angles, sayγ, is 90◦, the formula becomes
Cosine Law
By changing the side of the triangle fixed on the positive x-axis, we get:
Theorem (Cosine Law)
a2 =b2+c2−2bccosα
b2 =a2+c2 −2accosβ
c2 =a2+b2−2abcosγ
Note:
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 =
102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102
+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102
−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)
⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,
0◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =
120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
⇒α =β =
180◦−120◦
2 =30
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
⇒α =β = 180◦−2120◦ =
Example: Solve the triangle witha= 10, b = 10, and c= 10√3.
Solution:
(10√3)2 = 102+ 102−2(10)(10)(cosγ)⇒cosγ =−1 2,0
◦ < γ <
180◦
⇒γ =120◦
Since4ABC is isosceles,
Applications
Definition
E: observer’s eye, O: point being observed
line of sight – line joiningE and O
angle formed by the horizontal and the line of sight:
angle of elevation, ifO is above the horizontal throughE
Applications
Definition
E: observer’s eye, O: point being observed line of sight– line joining E and O
angle formed by the horizontal and the line of sight:
angle of elevation, ifO is above the horizontal throughE
Applications
Definition
E: observer’s eye, O: point being observed line of sight– line joining E and O
angle formed by the horizontal and the line of sight:
angle of elevation, ifO is above the horizontal throughE
Applications
Definition
E: observer’s eye, O: point being observed line of sight– line joining E and O
angle formed by the horizontal and the line of sight: angle of elevation, ifO is above the horizontal throughE
Applications
Definition
E: observer’s eye, O: point being observed line of sight– line joining E and O
angle formed by the horizontal and the line of sight: angle of elevation, ifO is above the horizontal throughE
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦
= a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦
= 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦
= 6b
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = b6
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = b6
⇒b= 6 tan 45◦
= 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = b6
⇒b= 6 tan 45◦ = 6(1) = 6
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = b6
⇒b= 6 tan 45◦ = 6(1) = 6
height
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = b6
⇒b= 6 tan 45◦ = 6(1) = 6
height =a+b
Roni, standing 6 ft away from a lamp post, observed that the angle of elevation to the top of the lamp post is30◦, while the angle of depression to the bottom of the lamp post is 45◦. How high is the lamp post?
Solution:
tan 30◦ = a6
⇒a= 6 tan 30◦ = 6
√ 3 3
=
2√3
tan 45◦ = b6
⇒b= 6 tan 45◦ = 6(1) = 6
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ =
h
a ⇒a= h
tan 60◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha
⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ =
h √ 3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ =
h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h a+1
⇒h=
√ 3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h
√
3 + 1) =
h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h= √
3 3 (
h √
3 + 1) =
h
3 +
√
3 3
Lana the Ant observed that the top of a jar of sugar has an angle of elevation of 30◦. After walking 1 ft towards the jar, the new angle of elevation is 60◦. What is the height of the jar?
Solution:
tan 60◦ = ha ⇒a= tan 60h ◦ = √h
3
tan 30◦ = h
a+1 ⇒h= √
3
3 (a+ 1)
⇒h=
√ 3 3 (
h √
3 + 1) = h 3 +
√ 3 3
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O:
N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E
path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course
75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O:
S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W
path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
Definition
O, P: pointsbearing or direction of P from O: measure of the acute angle whichOP makes with the north–south line
course (navigation): measure of the angle formed by the path of a plane or ship with the line due north, measured clockwise
First figure:
bearing of P from O: N75◦E path toP from O has course75◦.
Second figure:
bearing of P from O: S60◦W path toP from O has course
A ship traveled 10 km N30◦E. Then, it turned in the direction of S15◦E and traveled a distance of 20√2km. How far is the ship now from its origin?
Solution:
Letc be the distance of the ship from its origin.
c2 = 102 + (20√2)2
−2(10)(20√2) cos 45◦
c2 = 100 + 800
−2(10)(20√2)
√ 2 2
c2 = 900−400
c2 = 500
A ship traveled 10 km N30◦E. Then, it turned in the direction of S15◦E and traveled a distance of 20√2km. How far is the ship now from its origin?
Solution:
Letc be the distance of the ship from its origin.
c2 = 102 + (20√2)2
−2(10)(20√2) cos 45◦
c2 = 100 + 800
−2(10)(20√2)
√ 2 2
c2 = 900−400
c2 = 500
A ship traveled 10 km N30◦E. Then, it turned in the direction of S15◦E and traveled a distance of 20√2km. How far is the ship now from its origin?
Solution:
Letc be the distance of the ship from its origin.
c2 = 102 + (20√2)2
−2(10)(20√2) cos 45◦
c2 = 100 + 800
−2(10)(20√2)
√ 2 2
c2 = 900−400
c2 = 500
A ship traveled 10 km N30◦E. Then, it turned in the direction of S15◦E and traveled a distance of 20√2km. How far is the ship now from its origin?
Solution:
