Question

A person observes the top P of a vertical tower OP of height h from a station ${{S}_{1}}$ and finds that ${{\beta }_{1}}$ is the angle of elevation. He moves in a horizontal plane to second station ${{S}_{2}}$ and finds that $\angle P{{S}_{2}}{{S}_{1}}$ is ${{\gamma }_{1}}$ and the angle subtended by ${{S}_{2}}{{S}_{1}}$ at P is ${{\delta }_{1}}$ and the angle of elevation is ${{\beta }_{2}}$ . He moves again to a third station ${{S}_{3}}$ such that ${{S}_{3}}{{S}_{2}}={{S}_{2}}{{S}_{1}},\angle P{{S}_{3}}{{S}_{2}}={{\gamma }_{2}}$ and the angle subtended by ${{S}_{3}}{{S}_{2}}$ at P is ${{\delta }_{2}}$ . Show that \[\dfrac{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}{\sin {{\delta }_{1}}}=\dfrac{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}{\sin {{\delta }_{2}}}=\dfrac{h}{{{S}_{1}}{{S}_{2}}}\] .

Answer 1

Hint: Draw the suitable diagram with the help of given information’s provided and represent all the sides and angles given in the problem. Take $\sin {{\beta }_{1}}$ and $\sin {{\beta }_{2}}$ in right angle triangles $OP{{S}_{1}}$ and $OP{{S}_{2}}$ respectively. And use sine rule in the triangles $P{{S}_{1}}{{S}_{2}}$ and $P{{S}_{2}}{{S}_{3}}$ with the involvement of sides $P{{S}_{1}}$ and ${{S}_{2}}{{S}_{1}}$ for triangle $P{{S}_{1}}{{S}_{2}}$ and sides $P{{S}_{2}}$ and ${{S}_{3}}{{S}_{2}}$ for triangle $P{{S}_{2}}{{S}_{3}}$ . Sine rule in given as
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Where a, b, c are sides of a triangle with angles A, B, C lying opposite to sides a, b, c respectively. And use:
$\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotaneous}}$

Complete step-by-step solution -

Let us draw a suitable diagram with the help of the given information’s in the problem. So, information’s provided in the problem are:
(i) Vertical tower if of height ‘h’ (OP) and person is observing the top point P from a station ${{S}_{1}}$ with angle of elevation as ${{\beta }_{1}}$
(ii) Station ${{S}_{2}}$ is lying towards the tower from station ${{S}_{1}}$ , with $\angle P{{S}_{2}}{{S}_{1}}={{\gamma }_{1}}$ and angle subtended by ${{S}_{2}}{{S}_{1}}$ at P is ${{\delta }_{1}}$ and angle of elevation is ${{\beta }_{2}}$ .
(iii) Person moves from ${{S}_{2}}$ to ${{S}_{3}}$ such that ${{S}_{3}}{{S}_{2}}={{S}_{2}}{{S}_{1}}$ and $\angle P{{S}_{3}}{{S}_{2}}={{\gamma }_{2}}$ and angle subtended at P is ${{\delta }_{2}}$ .
And hence, we need to show the result with the above information’s. So, we have to prove:
$\dfrac{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}{\sin {{\delta }_{1}}}=\dfrac{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}{\sin {{\delta }_{2}}}=\dfrac{h}{{{S}_{1}}{{S}_{2}}}$
So, diagram can be represented as
Where, it is also given that
${{S}_{3}}{{S}_{2}}={{S}_{2}}{{S}_{1}}..................\left( i \right)$
Now, as we know $\sin \theta $ for any right-angle triangle is defined as
$\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotaneous}}...............\left( ii \right)$
So, in $\Delta OP{{S}_{1}}$ , we get
$\sin {{\beta }_{1}}=\dfrac{h}{P{{S}_{1}}}$
On cross-multiplying the above equation, we get
$P{{S}_{1}}=\dfrac{h}{\sin {{\beta }_{1}}}....................\left( iii \right)$
Similarly, in \[\Delta OP{{S}_{2}}\] , we can use equation (ii) and hence, we get
$\sin {{\beta }_{2}}=\dfrac{h}{P{{S}_{2}}}$
On cross-multiplying, the above equation, we get
$P{{S}_{2}}=\dfrac{h}{\sin {{\beta }_{2}}}..................\left( iv \right)$
As, we know sine rule for any triangle is given as
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}.................\left( v \right)$
Where a, b, c are sides of triangle with angles A, B, C which are lying just opposite to the sides a, b, c respectively.
Now, applying the sine rule in $\Delta P{{S}_{1}}{{S}_{2}}$ , we get
$\dfrac{\sin {{\delta }_{1}}}{{{S}_{2}}{{S}_{1}}}=\dfrac{\sin {{\gamma }_{1}}}{P{{S}_{1}}}=\dfrac{\sin {{\beta }_{1}}}{P{{S}_{2}}}$
Taking only first two terms of the above equation, we get
$\dfrac{\sin {{\delta }_{1}}}{{{S}_{2}}{{S}_{1}}}=\dfrac{\sin {{\gamma }_{1}}}{P{{S}_{1}}}..................\left( vi \right)$
Similarly, applying the sine rule with $\Delta P{{S}_{2}}{{S}_{3}}$ , we get
$\dfrac{\sin {{\delta }_{2}}}{{{S}_{3}}{{S}_{2}}}=\dfrac{\sin {{\gamma }_{2}}}{P{{S}_{2}}}=\dfrac{\sin {{\beta }_{2}}}{P{{S}_{3}}}$
Taking first two terms of the above equation, we get
$\dfrac{\sin {{\delta }_{2}}}{{{S}_{3}}{{S}_{2}}}=\dfrac{\sin {{\gamma }_{2}}}{P{{S}_{2}}}...................\left( vii \right)$
Now, putting value of $P{{S}_{1}}$ from equation (iii) to equation (vi), we get
$\begin{align}
  & \dfrac{\sin {{\delta }_{1}}}{{{S}_{2}}{{S}_{1}}}=\dfrac{\sin {{\gamma }_{1}}}{\left( \dfrac{h}{\sin {{\beta }_{1}}} \right)} \\
 & \dfrac{\sin {{\delta }_{1}}}{{{S}_{2}}{{S}_{1}}}=\dfrac{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}{h} \\
 & \Rightarrow \dfrac{\sin {{\delta }_{1}}}{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}=\dfrac{{{S}_{2}}{{S}_{1}}}{h}.................\left( viii \right) \\
\end{align}$
Similarly, putting value of $P{{S}_{2}}$ from the equation (iv) to the equation (vii), we get
$\begin{align}
  & \dfrac{\sin {{\delta }_{2}}}{{{S}_{3}}{{S}_{2}}}=\dfrac{\sin {{\gamma }_{2}}}{\left( \dfrac{h}{\sin {{\beta }_{2}}} \right)} \\
 & \dfrac{\sin {{\delta }_{2}}}{{{S}_{3}}{{S}_{2}}}=\dfrac{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}{h} \\
 & \Rightarrow \dfrac{\sin {{\delta }_{2}}}{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}=\dfrac{{{S}_{3}}{{S}_{2}}}{h}.....................\left( ix \right) \\
\end{align}$
Now, as we know ${{S}_{3}}{{S}_{2}}={{S}_{2}}{{S}_{1}}$ from the equation (i), so we can write the equation (ix) by replacing ${{S}_{3}}{{S}_{2}}$ by ${{S}_{2}}{{S}_{1}}$ .
So, we get equation (ix) as
$\dfrac{\sin {{\delta }_{2}}}{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}=\dfrac{{{S}_{2}}{{S}_{1}}}{h}.................\left( x \right)$
Now, comparing equation (viii) and (x), we get
$\dfrac{\sin {{\delta }_{1}}}{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}=\dfrac{\sin {{\delta }_{2}}}{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}=\dfrac{{{S}_{2}}{{S}_{1}}}{h}$
On reversing the fractions involved in the above equation, we get
$\dfrac{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}{\sin {{\delta }_{1}}}=\dfrac{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}{\sin {{\delta }_{2}}}=\dfrac{h}{{{S}_{2}}{{S}_{1}}}$
$\Rightarrow \dfrac{\sin {{\gamma }_{1}}\sin {{\beta }_{1}}}{\sin {{\delta }_{1}}}=\dfrac{\sin {{\gamma }_{2}}\sin {{\beta }_{2}}}{\sin {{\delta }_{2}}}=\dfrac{h}{{{S}_{1}}{{S}_{2}}}$
Hence the expression in the problem is proved.

Note: One may go wrong if he/she draws the diagram as

As $P{{S}_{2}}{{S}_{1}}$ can not be equal to the angle of elevation. So, as it is not mentioned in the question, that in which order ${{S}_{1}},{{S}_{2}},{{S}_{3}}$ will lie, we need to understand it with the help of given information in the problem. Hence, one needs to take care of this part while drawing the diagram.
One may use any other trigonometric functions in the triangles formed in the figure, which will make the solution complex and one may not get the given result. So, as the given expression has the involvement of only sin function, so, using sin function within the triangles formed in the diagram is the key point of the question.
One may go wrong with the positions of terms involved in the sine rule. So, be clear with the positions of sides and angles sin the formula of sine rule.