Angle Of Elevation And Depression

Q.1 You are standing 20 feet away from a tree, and you measure the angle of elevation to be 38°. How tall is the tree?

Solution:

The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume that you are 5 feet tall. Then the figure below shows the triangle you are solving.

The figure shows us that once we find the value of T, we have to add 5 feet to this value to find the total height of the triangle. To find T, we should use the tangent value:

tan ? ( 38 ? ) = {\displaystyle \tan(38^{\circ })=\,\!}	opposite adjacent = T 20 {\displaystyle {\frac {\text{opposite}}{\text{adjacent}}}={\frac {T}{20}}}
tan ? ( 38 ? ) = {\displaystyle \tan(38^{\circ })=\,\!}	T 20 {\displaystyle {\frac {T}{20}}}
T = {\displaystyle T=\,\!}	20 tan ? ( 38 ? ) ≈ 15.63 {\displaystyle 20\tan(38^{\circ })\approx 15.63\,\!}
Height of tree ≈ {\displaystyle {\text{Height of tree}}\approx \,\!}	20.63   ft {\displaystyle 20.63\ {\text{ft}}\,\!}

Q.2 You are standing on top of a building, looking at park in the distance. The angle of depression is 53°. If the building you are standing on is 100 feet tall, how far away is the park? Does your height matter?

Solution:

If we ignore the height of the person, we solve the following triangle:

Given the angle of depression is 53°, angle A in the figure above is 37°. We can use the tangent function to find the distance from the building to the park:

tan ? ( 37 ? ) = {\displaystyle \tan(37^{\circ })=\,\!}	opposite adjacent = d 100 {\displaystyle {\frac {\text{opposite}}{\text{adjacent}}}={\frac {d}{100}}}
tan ? ( 37 ? ) = {\displaystyle \tan(37^{\circ })=\,\!}	d 100 {\displaystyle {\frac {d}{100}}}
d = {\displaystyle d=\,\!}	100 tan ? ( 37 ? ) ≈ 75.36   ft {\displaystyle 100\tan(37^{\circ })\approx 75.36\ {\text{ft}}\,\!}

If we take into account the height if the person, this will change the value of the adjacent side. For example, if the person is 5 feet tall, we have a different triangle:

tan ? ( 37 ? ) = {\displaystyle \tan(37^{\circ })=\,\!}	opposite adjacent = d 105 {\displaystyle {\frac {\text{opposite}}{\text{adjacent}}}={\frac {d}{105}}}
tan ? ( 37 ? ) = {\displaystyle \tan(37^{\circ })=\,\!}	d 105 {\displaystyle {\frac {d}{105}}}
d = {\displaystyle d=\,\!}	105 tan ? ( 37 ? ) ≈ 79.12   ft {\displaystyle 105\tan(37^{\circ })\approx 79.12\ {\text{ft}}\,\!}