PROBLEMS ON ANGLE OF ELEVATION AND DEPRESSION

Problem 1 :

A surveyor is standing 100 meters from a bridge. She determines that the angle of elevation to the top of the bridge is 35°. If her height is 2.5 meters tall, then what is the height of the bridge?

Solution:

answer-q1.png

tan(35°) = x/100

100 tan 35° = x

100(0.7002) = x

x = 70.02 m

Since,  If her height is 2.5 meters tall creating the horizontal distance, so add the 2.5 m to the 70.02 m to determine the total height of the bridge.

70.02 m + 2.5 m = 72.52 m

The height of the bridge is 72.52 m.

Problem 2 :

From the top of a lighthouse, the angle of depression to a buoy is 25°. If the top of the lighthouse is 150 feet above sea level, find the horizontal distance from the buoy to the lighthouse.

Solution:

answer-q2
tan 25°=150xx=150tan 25°x=1500.466x=321.8 feet

Problem 3 :

A jet takes off and travels 56 miles to achieve an altitude of 12.6 miles. What is the angle of elevation for the jets path?

Solution:

answer-q3
sin 𝜃=12.656𝜃=sin-112.656𝜃=13°

Problem 4 :

Avery is at the top of the Eiffel tower. When looking out she sees the Seine River at an angle of depression of 62°. The height of the tower is 324 meters. What is the horizontal distance from the Eiffel Tower to the River.

Solution:

answer4

Let x be the horizontal distance from the Eiffel Tower to the river.

tan 62°=324xx=324tan 62°x=3241.880x=172.276 m

So, the horizontal distance from the Eiffel Tower to the river is 172.28 m.

Problem 5 :

Find ''x''.

answer-q5

Solution:

sin 25°=5xx=5sin 25°x=50.422x=11.84

Problem 6 :

How tall is a tree if it casts a shadow of 50 feet and they rays of the sun meet the ground at a 25° angle?

Solution:

answer-q6

Let x be the tall of the tree,

tan 25°=x50x=tan 25°×50x=0.466×50x=23.32 feet

So, the tall of the tree is 23.32 feet.

Problem 7 :

A sonar operator on a battleship detects a submarine at a distance of 500 m (horizontally) and at an angle of depression of 37°. How deep is the submarine?

Solution:

answer-q7
sin 37°=x500x=sin 37°×500x=0.601×500x=300.9 m

So, submarine is 300.9 meters deep.

Problem 8 :

Find the measure of the indicated angle.

answer-q8.png

Solution:

tan x = 615x=tan-1615x=21.801°

Problem 9 :

The angle of elevation of a 110 foot crane is 45°. How high can the crane raise building material?

answer-q9.png

Solution:

sin 45°=x110x=sin 45°×110x=0.707×110x=77.78 ft

Problem 10 :

Find ''x''.

answer-q10

Solution:

sin 70°=x3x=sin 70°×3x=0.939×3x=2.819

Problem 11 :

A boy flying a kite lets out 100 feet of string making an angle of elevation of 40°. How high above the ground is the kite?

Solution:

answer-q11
sin 40°=x100x=sin 40°×100x=0.642×100x=64.27 ft

So, the kite above the ground is 64.27 ft.

Problem 12 :

As viewed from a cliff 360 m above sea level, the angle of depression to a ship is 28°. How far is the ship from the shore?

Solution:

answer-q12
tan 28°=360xx=360tan 28°x=3600.5317x=677 m

Problem 13 : 

The angle of elevation from a ship to the top of a lighthouse is 3°. If the ship is 1000 km from the lighthouse, how tall is the lighthouse?

Solution:

answer-q13
tan 3°=x1000x=tan 3°×1000x=0.052×1000x=52.40 km

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