# The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is $60^{o}$. From a point Y, $40\ m$ vertically above X, the angle of elevation of the top Q of tower is $45^{o}$. Find the height of the tower PQ and the distance PX. $( Use\ \sqrt{3} \ =\ 1.73)$

**Given: **The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is $60^{o}$. From a point Y, $40\ m$ vertically above X, the angle of elevation of the top Q of tower is $45^{o}$.

**To do: **To find the height of the tower PQ and the distance PX.

**Solution:**

$MP= YX=40\ m$

$\therefore QM = h –40$

In right angled $\vartriangle QMY$,

$tan45^{o} =\frac{QM}{MY} =1=\frac{h-40}{PX} \ \ \ \ \ \ \ \ ( \because \ MY\ =\ PX)$

$\therefore PX = h –40 \ \ \ \ \ \ ....( 1)$

In right angled $\vartriangle QPX$,

$tan\ 60^{o} =\frac{OP}{PX} =\sqrt{3} =\frac{h}{PX} =\frac{h}{h-40} \ \ \ ....( 2)$

$\Rightarrow \frac{h}{h-40} =\sqrt{3}$

$\Rightarrow h=h\sqrt{3} -40\sqrt{3}$

$\Rightarrow h\sqrt{3} -h=40\sqrt{3}$

$\Rightarrow h\times 1.73-h=40\times 1.73$

$\Rightarrow 0.73h=69.2$

Or $h=\frac{69.2}{0.73}$

Thus, $PQ$ is $94.79\ m$.

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