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

Question

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

Vedantu Content Team · Accepted Answer

Hint: In this question, we are supposed to find the length of tower and distance between point P and point X. We are going to consider the $\Delta YRQ$ and find the value of YR and XP from there and similarly, find the value of PX from it.
In $\Delta YRQ$, we have,
$\tan {45^ \circ } = \dfrac{{QR}}{{YR}}$
$1 = \dfrac{x}{{YR}}$
$ \Rightarrow YR = x$
Or $XP = x\left[ {As\,YR = XP} \right]$ …..(1)
Now, In $\Delta XPQ$ , we have,
$\tan {60^ \circ } = \dfrac{{PQ}}{{PX}}$
$\sqrt 3 = \dfrac{{x + 40}}{x}$ [Using (1)]
$\sqrt 3 x = x + 40$
$XQ = 109.3m$
$x = \dfrac{{40}}{{\sqrt 3 - 1}}$
On rationalising the denominator, we get,
$x = \dfrac{{40}}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}$
$x = \dfrac{{40\sqrt 3 + 1}}{{3 - 1}}$
$x = \dfrac{{40\sqrt 3 + 1}}{2}$
$x = 20\left( {\sqrt 3 + 1} \right) = 54.64m$
So, height of the tower, PQ= x+40 = 54.64 + 40 = 94.64m
Now in $\Delta XPQ$, we have,
$\sin {60^ \circ } = \dfrac{{PQ}}{{XQ}}$
$\dfrac{{\sqrt 3 }}{2} = \dfrac{{94.64}}{{XQ}}$
$XQ = \dfrac{{94.64 \times 2}}{{\sqrt 3 }}$
$XQ = \dfrac{{94.64 \times 2 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}$
Note: In these questions, make sure you take the correct trigonometric angle.
Answer = $XQ = 109.3m$

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

NCERT Book Solutions

Reference Book Solutions

Question Papers

Exams

Quick Links

Book your FREE Online counselling session

Your FREE Online counselling session has been booked!

Please try again later

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

NCERT Book Solutions

Reference Book Solutions

Question Papers

Exams

Quick Links