# A flagpole sits on the top of a building. Angles of elevation are measured from a point 500 feet away from the building. The angle of elevation to the top of the building is 38˚ and to the top of the flagpole 42˚. How high is the flagpole?

Apr 10, 2017

The height of the flagpole is approximately $60$ feet.

#### Explanation:

Always try to draw a diagram.

We know that there is a right angle between the ground and the building. Therefore, we can use the 3 basic trig ratios instead of the sine or cosine law to solve this problem.

Since the angle in the corner of the larger right triangle measures 42˚, the top angle in this triangle measures 180˚ - 90˚ - 42˚ = 48˚.

By basic trig ratios, we can find the height of the building with the flag pole on top, call it $H$.

(tan42˚)/1 = H/500

H = 500tan42˚

I would keep it in exact form until the last step.

We now devise an expression for the height of the building (without the flag pole). Call it $a$

(tan38˚)/1 = a/500

a = 500tan38˚

We can now state that

$h = H - a$

h = 500tan42˚ - 500tan38˚

$h \approx 59.559 \approx 60 \text{feet}$

Hopefully this helps!

Apr 10, 2017

The flagpole is $60 f t$ in height to the nearest foot.

#### Explanation:

1) The flagpole is on top of a building.
2)Angles of elevation both measured from point $500 f t$ from building
3) Angle of elevation to the top of building is $38 \mathrm{de} g$
4) Angle of elevation to the top of flagpole is $42 \mathrm{de} g$

The information above will provide us with two right angle triangles, one smaller one inside a larger one.

Both will have a base of $500 f t$.

The smaller triangle will have a base angle $\beta$ of $38$deg opposite the $90$deg, and the larger triangle will have a base angle $\beta$ of $42$deg.

From this information we can find the heights of the building and the building + pole using the definition of the tangent of the two base angles $\beta$:

$\tan \left(\beta\right) = \frac{o p p}{a \mathrm{dj}}$ where the $o p p$ is the height and the $a \mathrm{dj}$ is the $500 f t$

$o p p \left(b u i l d\right) = \tan \left(38\right) \cdot \left(500 f t\right) = 390.6 f t =$height of building

$o p p \left(f l a g\right) = \tan \left(42\right) \cdot \left(500 f t\right) = 450.2 f t =$height of building + pole

Then to the nearest foot the height of the flagpole is:

$450.2 f t - 390.6 f t = 60 f t$