In the following figure, a tower
𝐴𝐵 is 20 metres high, and its shadow on the ground 𝐵𝐶 is 20 root three metres
long. Find the angle of elevation of the
Sun from point 𝐶.
We’re told in the question that, in
the right-angled triangle 𝐴𝐵𝐶, the length of 𝐴𝐵 is 20 metres and the length of
𝐵𝐶 is 20 root three metres. In order to solve this problem,
we’ll use our trigonometrical ratios, also sometimes known as SOHCAHTOA.
Our first step is to label all
three sides. The length 𝐴𝐶 is the hypotenuse,
as it is the longest side and is opposite the right angle. The length 𝐴𝐵 is the opposite, as
it is opposite the angle 𝜃. And finally, the length 𝐵𝐶 is the
adjacent, as it is next to or adjacent to the angle 𝜃 and the 90-degree angle.
As we know the length of the
opposite and the adjacent, we will use the tan or tangent ratio. Tan 𝜃 is equal to the opposite
divided by the adjacent. Substituting in the values of 𝐴𝐵
and 𝐵𝐶 gives us tan 𝜃 is equal to 20 divided by 20 root three.
We can simplify this fraction by
dividing the numerator and denominator by 20. Remember, whatever you do to the
top you must do to the bottom. 20 divided by 20 is equal to one,
and 20 root three divided by 20 is equal to root three. We therefore have tan 𝜃 is equal
to one divided by root three or one over root three.
We would normally use our
calculator to solve the end of this problem. However, in this case, the answer
is one of our special trig angles, as tan 30 is equal to one over root three. This means that 𝜃 is equal to 30
degrees. We can therefore say that the angle
of elevation of the Sun from point 𝐶 is equal to 30 degrees. To show this on the diagram, we
could extend the hypotenuse up to the Sun at the same angle from point 𝐶.