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Question: A church tower stands on the bank of a river, which is 150 feet wide, and on the top of the tower is...

A church tower stands on the bank of a river, which is 150 feet wide, and on the top of the tower is a spire 30 feet high. To an observer on the opposite bank of the river, the Spiro subtends the same angle that a pole six feet high subtends when placed upright on the ground at the foot of the tower. Prove that the height of the tower is nearly 285 feet

Explanation

Solution

Hint : In this question distance of the church from the river bank is given and height of the spite is given so by applying the angle of elevation we will find the height of the church by comparing the two equations observed, one the height of church and other the total height of church and the spire.

Complete step-by-step answer :
Distance between church tower and the river bank =150ft = 150ft
Height of the spire=30ft = 30ft
Now let the angle subtended by the Spiro to an observer on the river bank of river be α\alpha
It is said that the angle subtended by the Spiro is same as the angle subtended by the pole which is at the ground as shown in the diagram below

From the figure
POQ=ROS=α\angle POQ = \angle ROS = \alpha {Same angle subtended by observer}
Now consider ΔROS\Delta ROS
Here
tanα=RSOS=6150=125(i)\tan \alpha = \dfrac{{RS}}{{OS}} = \dfrac{6}{{150}} = \dfrac{1}{{25}} - - (i)
Now let the height QS=h(ii)QS = h - - (ii)
Also QOR=θ\angle QOR = \theta
Hence we can say
POS=2α+θ(iii)\angle POS = 2\alpha + \theta - - (iii)
Now in ΔQOS\Delta QOS
tan(QOS)=h150(iv)\tan \left( {QOS} \right) = \dfrac{h}{{150}} - - (iv)
In ΔPOS\Delta POS
tan(POS)=h+30150(v)\tan \left( {POS} \right) = \dfrac{{h + 30}}{{150}} - - (v)
Now we can write
tan(2α+θ)=tan(α+α+θ)\tan \left( {2\alpha + \theta } \right) = \tan \left( {\alpha + \alpha + \theta } \right)
Hence by substituting the values from the equation (v) we can write

h+30150=h150+61501h1506150 h+30150=h+61506h  \Rightarrow \dfrac{{h + 30}}{{150}} = \dfrac{{\dfrac{h}{{150}} + \dfrac{6}{{150}}}}{{1 - \dfrac{h}{{150}} \cdot \dfrac{6}{{150}}}} \\\ \dfrac{{h + 30}}{{150}} = \dfrac{{h + 6}}{{150 - 6h}} \;

Now by cross multiplying, we get

150h6h2+4500180h=150h+900 6h2+180h5400=0 h2+30h900=0  \Rightarrow 150h - 6{h^2} + 4500 - 180h = 150h + 900 \\\ \Rightarrow 6{h^2} + 180h - 5400 = 0 \\\ \Rightarrow {h^2} + 30h - 900 = 0 \;

By solving the obtained quadratic equation we get the height

h=15(4011) =285ft  \Rightarrow h = 15\left( {\sqrt {401} - 1} \right) \\\ = 285ft \;

Hence, the height of the tower is nearly 285 feet

Note : Students should not be confused with the angle of elevation and the angle of depression. These are two different things. The angle of elevation is the rising angle generally, from the foot while the angle of depression is measured for the top of the tower.