PROBLEMS USING SOHCAHTOA

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In trigonometry, we have six trigonometric ratios.

sin Ɵ, cos Ɵ,  tan Ɵ, cosec Ɵ, sec Ɵ, cot Ɵ

sin Ɵ = Opposite side / Hypotenuse

cos Ɵ = Adjacent side / Hypotenuse

tan Ɵ = Opposite side / Adjacent side

cosec Ɵ = Hypotenuse / Opposite side

sec Ɵ = Hypotenuse / Adjacent side

cot Ɵ = Adjacent side / Opposite side

right-triangle-opposite-adjacent-hypotenuse

Using the phrase, SOHCOATOA we can remember the formulas easily.

Find the value of the trigonometric ratio. Express answers as a fraction in lowest terms.

Problem 1 :

Find sin x.

problems-on-sohcahtoa-q1

Solution :

sin x = opposite/hypotenuse

= AB/AC

= 3/5

So, sin x = 3/5.

Problem 2 :

Find cos x.

problems-on-sohcahtoa-q2.png

Solution :

cos x = adjacent/hypotenuse

= BC/AC

= 40/41

So, cos x = 40/41.

Problem 3 :

Find tan x

problems-on-sohcahtoa-q3.png

Solution :

tan x = opposite/adjacent

= AB/BC

= 8/6

= 4/3

So, tan x = 4/3.

Problem 4 :

Find cos x.

problems-on-sohcahtoa-q4.png

Solution :

cos x = adjacent/hypotenuse

= BC/AC

= 4/5

So, cos x = 4/5.

Problem 5 :

Find sin x.

problems-on-sohcahtoa-q5.png

Solution :

sin x = opposite/hypotenuse

= AB/AC

= 9/41

So, sin x = 9/41.

Problem 6 :

Find sin x

problems-on-sohcahtoa-q6.png

Solution :

sin x = opposite/hypotenuse

= 8/10

= 4/5

So, sin x = 4/5.

Problem 7 :

Find tan x.

problems-on-sohcahtoa-q7.png

Solution :

tan x = opposite/adjacent

= AB/BC

= 3/4

So, tan x = 3/4.

Problem 8 :

Find tan x.

problems-on-sohcahtoa-q8.png

Solution :

tan x = opposite/adjacent

= AB/BC

= 9/40

So, tan x = 9/40.

Problem 9 :

Fins cos x.

problems-on-sohcahtoa-q9.png

Solution :

cos x = adjacent/hypotenuse

= BC/AC

= 6/10

So, cos x = 3/5.

Problem 10 :

Find tan x.

problems-on-sohcahtoa-q10.png

Solution :

tan x = opposite/adjacent

= AB/BC

= 12/9

So, tan x = 4/3.

Problem 11 :

Find cos x.

problems-on-sohcahtoa-q11.png

Solution :

cos x = adjacent/hypotenuse

= BC/AC

= 9/15

= 3/5

So, cos x = 3/5.

Problem 12 :

Find sin x

problems-on-sohcahtoa-q12.png

Solution :

sin x = opposite/hypotenuse

= AB/AC

= 12/15

= 4/5

So, sin x = 4/5.

Problem 13 :

You are hiking near a canyon. While standing at A, you measure an angle of 90º between B and C, as shown. You then walk to B and measure an angle of 76° between A and C. The distance between A and B is about 2 miles. How wide is the canyon between A and C?

problems-on-sohcahtoa-q13.png

Solution :

tan x = opposite side / adjacent side

Opposite side = AC, Hypotenuse = BC and adjacent side = AB

tan 76 = AC/AB

4 = x/2

x = 4(2)

x = 8

Problem 14 :

A parasailer is attached to a boat with a rope that is 72 feet long. The angle of elevation from the boat to the parasailer is 28°. Estimate the parasailer’s height above the boat.

Solution :

By observing the given information, 

problems-on-sohcahtoa-q14.png

sin x = opposite side / hypotenuse

opposite side = h, hypotenuse = 72 ft

sin 28 = h/72

0.469 = h/72

h = 0.469(72)

= 33.76

So, the missing side is 33.76 ft

Problem 15 :

To measure the width of a river, you plant a stake on one side of the river, directly across from a boulder. You then walk 100 meters to the right of the stake and measure a 79° angle between the stake and the boulder. What is the width w of the river?

problems-on-sohcahtoa-q15.png

Solution :

tan x = opposite side / adjacent side

opposite side = w, adjacent side = 100 m

tan 79 = w/100

5.144 = w/100

w = 5.144(100)

w = 514.4

So, the missing side is 514.4 meter.

Problem 16 :

Katoomba Scenic Railway in Australia is the steepest railway in the world. The railway makes an angle of about 52° with the ground. The railway extends horizontally about 458 feet. What is the height of the railway at its highest point?

Solution :

problems-on-sohcahtoa-q16.png

Opposite side = AB = height of the railway

Adjacent side = BC = 458 ft

tan x = opposite side / adjacent side

tan 52 = AB/BC

1.2799 = AB/458

AB = 1.27(458)

= 581.6

Approximately 582 ft

Problem 17 :

A person whose eye level is 1.5 meters above the ground is standing 75 meters from the base of the Jin Mao Building in Shanghai, China. The person estimates the angle of elevation to the top of the building is about 80°. What is the approximate height of the building?

Solution :

problems-on-sohcahtoa-q17.png

CD = 1.5 meters

BC = 75 meters

AB = h

tan x = opposite side / adjacent side

tan 80 = AB/BC

5.67 = AB/75

AB = 5.67(75)

AB = 425.25

AE = AB + BE

= 425.25 + 1.5

= 426.75 meters

So, the required height is 426.75 meters.

Problem 18 :

You are standing on the Grand View Terrace viewing platform at Mount Rushmore, 1000 feet from the base of the monument.

problems-on-sohcahtoa-q18.png

a. You look up at the top of Mount Rushmore at an angle of 24°. How high is the top of the monument from where you are standing? Assume your eye level is 5.5 feet above the platform.

b. The elevation of the Grand View Terrace is 5280 feet. Use your answer in part (a) to find the elevation of the top of Mount Rushmore.

Solution :

a) tan x = opposite side / adjacent side

tan 24 = b/1000

0.445 = b/1000

b = 0.445(1000)

b = 445.2

eye level = 5.5 ft

above the platform, the required height = 445.2 + 5.5

= 450.7 ft

Approximately 451 ft

b) The elevation of Grand view Terrace = 5280 ft

= 5280 + 451

= 5731 ft

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