Question

A vertical tower of height 50 feet sands on a sloping ground. The foot of the tower is at the same level as the middle point of a vertical flagpole. The angles of depression from the top of the tower to the top and bottom of the pole are $15{}^\circ $ and $45{}^\circ $. Find the length of the pole.

Answer 1

Hint: Assume that the height of the pole is h. Use the fact that in a right angled triangle, tangent of an angle is equal to the ratio of the opposite side to the adjacent side in triangle ADG and hence prove that $GD=AG\tan 45{}^\circ $ and similarly in triangle DBF, prove that $DF=BF\tan 15{}^\circ $. Hence find the length h using h = GD-DF. Use the fact that DF+FC = 50 and hence form an equation in BF. Solve for BF and hence find the length h. Verify your answer.

Complete step by step answer:

Given: AB is a pole and DC is a tower whose foot C is at the mid-level of the tower AB. The angles of depression from the top of the tower to the top and bottom of the pole are $15{}^\circ $ and $45{}^\circ $ respectively. DC = 50 feet.
To determine: The height of the pole AB.
Let h be the height of the pole AB. Now, we have
$\angle DBF=15{}^\circ ,\angle DAG=45{}^\circ $
In triangle ADG, we have DG is the side opposite to angle A and AG is the side adjacent to angle A.
We know that in a right-angled triangle, the tangent of an angle is equal to the ratio of the opposite side to the adjacent side. Hence, we have
$\tan 45{}^\circ =\dfrac{DG}{AG}$
Multiplying both sides by AG, we get
$DG=AG\tan 45{}^\circ $
We know that AGFB is a rectangle. Hence, we have AG = BF.
Hence, we have
$DG=BF\tan 45{}^\circ \text{ }\left( i \right)$
In triangle DBF, we have DF is the side opposite to angle B and BF is the side adjacent to angle B.
We know that in a right-angled triangle, the tangent of an angle is equal to the ratio of the opposite side to the adjacent side. Hence, we have
$\tan 15{}^\circ =\dfrac{DF}{BF}$
Multiplying both sides by BF, we get
$DF=BF\tan 15{}^\circ \text{ }\left( ii \right)$
Subtracting equation (ii) from equation (i), we get
$GD-DF=BF\left( \tan 45{}^\circ -\tan 15{}^\circ \right)$
Hence, we have
$h=BF\left( \tan 45{}^\circ -\tan 15{}^\circ \right)\text{ }\left( iii \right)$
Also, we have
DC = 50
Hence, we have
$\begin{align}
  & DF+FC=50 \\
 & \Rightarrow DF+\dfrac{h}{2}=50 \\
\end{align}$
Substituting the values of DF and h, we get
$\begin{align}
  & BF\tan 15{}^\circ +\dfrac{BF\left( \tan 45{}^\circ -\tan 15{}^\circ \right)}{2}=50 \\
 & \Rightarrow BF\left( \tan 15{}^\circ +\tan 45{}^\circ \right)=100 \\
\end{align}$
Dividing both sides by $\tan 15{}^\circ +\tan 45{}^\circ $, we get
$BF=\dfrac{100}{\tan 45{}^\circ +\tan 15{}^\circ }$
Substituting the value of BF in equation (iii), we get
$h=\dfrac{100\left( \tan 45{}^\circ -\tan 15{}^\circ \right)}{\tan 45{}^\circ +\tan 15{}^\circ }\text{ }\left( iv \right)$
We know that $\tan x=\dfrac{\sin x}{\cos x}$
Hence, we have
$h=\dfrac{100\left( \dfrac{\sin 45{}^\circ }{\cos 45{}^\circ }-\dfrac{\sin 15{}^\circ }{\cos 15{}^\circ } \right)}{\dfrac{\sin 45{}^\circ }{\cos 45{}^\circ }+\dfrac{\sin 15{}^\circ }{\cos 15{}^\circ }}$
Multiplying numerator and denominator by $\cos 45{}^\circ \cos 15{}^\circ $, we get
$h=\dfrac{100\left( \sin 45{}^\circ \cos 15{}^\circ -\cos 45{}^\circ \sin 15{}^\circ \right)}{\sin 45{}^\circ \cos 15{}^\circ +\cos 45{}^\circ \sin 15{}^\circ }$
We know that sin(A-B) = sinAcosB-cosAsinB and sin(A+B) = sinAcosB+cosAsinB.
Hence, we have
$h=\dfrac{100\left( \sin \left( 45{}^\circ -15{}^\circ \right) \right)}{\sin \left( 45{}^\circ +15{}^\circ \right)}=\dfrac{100\sin 30{}^\circ }{\sin 60{}^\circ }$
We know that $\sin \left( 30{}^\circ \right)=\dfrac{1}{2}$ and $\sin \left( 60{}^\circ \right)=\dfrac{\sqrt{3}}{2}$

Hence, we have $h=\dfrac{100\left( \dfrac{1}{2} \right)}{\dfrac{\sqrt{3}}{2}}=\dfrac{100}{\sqrt{3}}$ feet which is the required height of the pole.

Note: [1] We can find the value of $\tan 15{}^\circ $ directly by observing that $15{}^\circ =45{}^\circ -30{}^\circ $
Hence taking tangents on both sides and using $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$, we get
$\tan 15{}^\circ =\dfrac{\tan 45{}^\circ -\tan 30{}^\circ }{\tan 45{}^\circ +\tan 30{}^\circ }=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$ and hence on substituting the value of $\tan 15{}^\circ $ in equation (iv), we get
$h=100\times \dfrac{1-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}{1+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}=100\times \dfrac{\sqrt{3}+1-\sqrt{3}+2}{\sqrt{3}+1+\sqrt{3}-1}=\dfrac{100}{\sqrt{3}}$