involving angles of elevation and depression.

### Solved Examples

**Question 1: **A surveyor wanted to measure the height of a mountain. He traveled in the straight road leading to the mountain and measured the angle of elevation of the peak from a point A which is on level ground with the mountain base as 32º. He drove 1 Km further down the road and again measured the angle of elevation of the peak as 40º. Find the height of the mountain nearest to a meter.

** Solution: **

As we see in the pictures, two variables h and x are introduced, the
height of the mountain and the distance BC. Using trig formulas we can
eliminate x and solve for h.

In right triangle BCD,

tan 40 = $\frac{h}{x}$ ⇒ x = $\frac{h}{tan 40}$

in right triangle ACD,

tan 32 = $\frac{h}{x+1}$

= $\frac{h}{\frac{h}{tan40}+1}$ Substitution

= $\frac{h\tan40}{h+tan40}$ Complex fraction simplified.

(h + tan 40) tan 32 = htan 40 Cross multiplication

htan 32 + tan 40.tan 32 = htan40

htan 40 - htan 32 = tan 40.tan 32

h(tan 40 - tan 32) = tan 40.tan 32

h = $\frac{tan40tan32}{tan40-tan32}$

≈ 2.4475 Km

Hence the estimated height of the mountain = 2,4475 meters.

**Question 2: **Roger views from his shop window which is 20 ft above ground level, the
top and the base of a building on the opposite side of the Road. If the
angles of elevation and depression of the top and base are 64º and 28º
find the height of the building he is viewing rounded to the tenth of a
foot.

The situation can be sketched as follows:

** Solution: **

AB represents the height at which Roger is viewing and = 20 ft. CD
represents the building he is viewing and we need to find its measure
which is equal to (20 + y) ft.

In right triangle AEC,

tan 28º = $\frac{CE}{AE}$ = $\frac{20}{x}$

x = $\frac{20}{tan28}$ ≈ 37.61 ft

In right triangle AED,

tan 64º = $\frac{y}{x}$ = $\frac{y}{37.61}$

y = 37.61 x tan 64 ≈ 77.12 ft.

Height of the building = 77.1 + 20 = 97.1 ft (rounded to the tenth of a foot).

**Question 3: **The shadow of a tower, when the angle of elevation of the Sun is 30º is
found to be 40 meters longer than when it is 45º. Find the height of the
Tower.

** Solution: **

The above sketch depicts the situation.

The height of the tower is represented by AB which is taken as y ft.

ABC is an isosceles right triangle ⇒ y = x.

In right triangle ABD,

tan 30º = $\frac{y}{x+40}$ = $\frac{x}{x+40}$ Definition of tangent of an angle.

x = (x + 40)tan 30 Cross multiplication.

x = x tan 30 + 40 tan 30

x- x tan 30 = 40 tan 30

x = $\frac{40tan30}{1-tan30}$ ≈ 54.6 ft.