Dedicated to Ayush Rai

Relevant Figure

Let the point from where he first observes the tower and the flagstaff be $A$ . Let the tower be represented by $PQ$ and the flagstaff by $PR$ . Also, let the second point of observation be $B$ .

Then, $\alpha = \angle PAR = \angle PBR \quad (\text{Given}) \\ \beta = \angle PAQ \quad (\text{Given}) \\ a = AB \quad (\text{Given}) \\ \theta = \angle BPA \quad (\text{Say})$

Note that the quadrilateral $ABPR$ is concylic (this follows from the first equality).

Hence the following equalities hold: $\beta = \angle PAQ = \angle BRQ \\ \theta = \angle BPA = \angle ARB$

Now, by the exterior angle property, $\angle APR = \angle AQP + \angle PAQ = 90^\circ + \beta$

As the sum of angles in $\Delta APR$ is $180^\circ$ , $\theta = 90^\circ - (\alpha + 2\beta)$

Applying the sine rule to appropriate triangles, we get $\frac{AP}{\sin (\theta + \beta)} = \frac{a}{\sin \theta} = \frac{PR}{\sin \alpha}$

$\therefore \text{ Height of tower }= PQ = AP \sin \beta = a \frac{ \sin \beta \cos (\alpha + \beta)}{\cos (\alpha+2\beta)} \\ \therefore \text{ Height of flagstaff } = PR = a \frac{ \sin \alpha}{\cos (\alpha+2\beta)}$

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Note: This problem and its solution have been taken from this book .

In the solution, instead of hit and trial we can use quadratic equation too.