Question

page 184: word problems : angle of elevation
in this diagram, ( x^circ ) marks the angle of elevation of the top of the tree as seen from a point on the ground.
the angle of elevation is always measured from the ground up. it is an upward angle from a horizontal line. it is always inside the triangle.
you can think of the angle of elevation in relation to the movement of your eyes. you are looking straight ahead and you must raise (elevate) your eyes to see the top of a tree.
when trying to remember the meaning of an angle of elevation think of an elevator that only goes up!
a radio station tower was built in two sections. from a point 87 feet from the base of the tower, the angle of elevation of the top of the first section is ( 25^circ ), and the angle of elevation of the top of the second section is ( 40^circ ). to the nearest foot, what is the height of the top section of the tower?
a 1.85m tall surveyor looks up at a building that is 30metres away. the angle of elevation is 77 degrees. how tall is this building?

Explanation:

First Problem (Radio Station Tower)

Step1: Find height of first section

We use the tangent function for the first section. Let \( h_1 \) be the height of the first section. The angle of elevation is \( 25^\circ \) and the adjacent side (distance from base) is 87 feet. So, \( \tan(25^\circ)=\frac{h_1}{87} \), which gives \( h_1 = 87\times\tan(25^\circ) \). Calculating, \( \tan(25^\circ)\approx0.4663 \), so \( h_1\approx87\times0.4663\approx40.57 \) feet.

Step2: Find height of second section

Let \( h_2 \) be the height of the second section (total height up to top of second section). The angle of elevation is \( 40^\circ \), so \( \tan(40^\circ)=\frac{h_2}{87} \). \( \tan(40^\circ)\approx0.8391 \), so \( h_2 = 87\times0.8391\approx73.00 \) feet.

Step3: Find height of top section

The height of the top section \( x = h_2 - h_1 \). Substituting values, \( x\approx73.00 - 40.57\approx32.43 \), which rounds to 32 feet.

Step1: Find height above surveyor

Let \( h \) be the height above the surveyor. The angle of elevation is \( 77^\circ \), distance is 30 meters. Using \( \tan(77^\circ)=\frac{h}{30} \). \( \tan(77^\circ)\approx4.3315 \), so \( h = 30\times4.3315\approx129.945 \) meters.

Step2: Find total building height

Add the surveyor's height (1.85 m) to \( h \). So total height \( = 129.945 + 1.85\approx131.795 \), which rounds to 132 meters.

Answer:

32 feet

Second Problem (Building Height)