SOLVING RIGHT TRIANGLES TRIGONOMETRY

For each triangle, write sin ϴ, cos ϴ and tan ϴ as fractions.

Problem 1 :

Solution :

From the triangle above,

Hypotenuse = 5

Opposite side = 3

Adjacent side = 4

Finding the value of sin ϴ :

sin ϴ = Opposite side / Hypotenuse

sin ϴ = 3/5

Finding the value of cos ϴ :

cos ϴ = Adjacent side / Hypotenuse

cos ϴ = 4/5

Finding the value of tan ϴ :

tan ϴ = Opposite side / Adjacent side

tan ϴ = 3/4

Problem 2 :

Solution :

From the triangle above,

Hypotenuse = 13

Opposite side = 5

Adjacent side = 12

Finding the value of sin ϴ :

sin ϴ = Opposite side / Hypotenuse

sin ϴ = 5/13

Finding the value of cos ϴ :

cos ϴ = Adjacent side / Hypotenuse

cos ϴ = 12/13

Finding the value of tan ϴ :

tan ϴ = Opposite side / Adjacent side

tan ϴ = 5/12

Problem 3 :

Solution :

From the triangle above,

Hypotenuse = 17

Opposite side = 15

Adjacent side = 8

Finding the value of sin ϴ :

sin ϴ = Opposite side / Hypotenuse

sin ϴ = 15/17

Finding the value of cos ϴ :

cos ϴ = Adjacent side / Hypotenuse

cos ϴ = 8/17

Finding the value of tan ϴ :

tan ϴ = Opposite side / Adjacent side

tan ϴ = 15/8

Problem 4 :

Solution :

From the triangle above,

Hypotenuse = 2.5

Opposite side = 2

Adjacent side = 1.5

Finding the value of sin ϴ :

sin ϴ = Opposite side / Hypotenuse

sin ϴ = 2/2.5

Finding the value of cos ϴ :

cos ϴ = Adjacent side / Hypotenuse

cos ϴ = 1.5/2.5

Finding the value of tan ϴ :

tan ϴ = Opposite side / Adjacent side

tan ϴ = 2/1.5

Problem 5 :

Solution :

From the triangle above,

Hypotenuse = 50

Opposite side = 48

Adjacent side = 14

Finding the value of sin ϴ :

sin ϴ = Opposite side / Hypotenuse

sin ϴ = 48/50

sin ϴ = 24/25

Finding the value of cos ϴ :

cos ϴ = Adjacent side / Hypotenuse

cos ϴ = 14/50

cos ϴ = 7/25

Finding the value of tan ϴ :

tan ϴ = Opposite side / Adjacent side

tan ϴ = 48/14

tan ϴ = 24/7

Problem 6 :

Solution :

Hypotenuse = 12.5

Opposite side = 3.5

Adjacent side = 12

Finding the value of sin ϴ :

sin ϴ = Opposite side / Hypotenuse

sin ϴ = 3.5/12.5

Finding the value of cos ϴ :

cos ϴ = Adjacent side / Hypotenuse

cos ϴ = 12/12.5

Finding the value of tan ϴ :

tan ϴ = Opposite side / Adjacent side

tan ϴ = 3.5/12

Problem 7 :

A passenger in an airplane sees two towns directly to the left of the plane.

a. What is the distance d from the airplane to the first town?

b. What is the horizontal distance x from the airplane to the first town?

c. What is the distance y between the two towns? Explain the process you used to find your answer.

Solution :

a) In the smaller triangle, the angle of elevation is 25 degree.

sin 25 = 25000/d

0.422 = 25000/d

d = 25000/0.422

= 59241.7

Approximately 59242 miles.

b) cos 25 = x/d

0.906 = x/59242

x = 0.906 (59242)

= 53691.4

Approximately 53691 miles

c) Distance between two towns :

In the large triangle, angle of elevation is 15 degree.

tan 15 = 25000/(x + y)

0.267 = 25000 / (53691 + y)

53691 + y = 25000/0.267

= 93632.9

Approximately 93633 miles

53691 + y = 93633

y = 93633 - 53691

= 39942 miles.

So, the required distance between two towns is 39942 miles.

Problem 8 :

You measure the angle of elevation from the ground to the top of a building as 32°. When you move 50 meters closer to the building, the angle of elevation is 53°. What is the height of the building?

Solution :

In triangle ABC,

tan 53 = AB/BC

AB = BC(tan 53)

= BC(1.327)

AB = 1.327 BC -------(1)

In triangle ABD,

tan 32 = AB/BD

AB = BD(tan 32)

= BD(0.624)

= 0.624 BD

AB = 0.624(BC + CD) -------(2)

(1) = (2)

1.327 BC = 0.624(BC + CD)

1.327 BC = 0.624 BC + 0.624 CD

1.327 BC - 0.624 BC = 0.624 (50)

0.703BC = 31.2

BC = 31.2/0.703

BC = 44.38

By applying the value of BC in (1), we get

AB = 1.327(44.38)

= 58.89

Approximately 59 meters.