Question

from a point on the ground the anglle of elevation of top and bottom of the tower is fixed at top of 10 m high are 30degree
and 45degree find the height of the tower​

Answer 1

Correct Question:-

From a point on the ground the angle of elevation of the bottom and top of the tower fixed at the top of a building 10m high are 30° and 45° respectively. Find the height of the tower.

Given:-

• Angle of elevation of the top and bottom of the tower from ground = 30° and 45° respectively.
• Height of the building = 10m

Note:-

Refer to the attachment for the diagram.

Assumption:-

Let the height of the tower be x

Solution:-

Here,

For the building we have to take right ∆ABC where,

BC = 10m

∠CAB = 30°

Therefore,

In ∆ABC

$\sf{Tan30^\circ = \dfrac{BC}{AB}}$

= $\sf{\dfrac{1}{\sqrt{3}} = \dfrac{10}{AB}}$

=> $\sf{AB = 10\sqrt{3}\:m}$

Now,

In ∆ABD,

AB = 10 m

BD = BC + CD = $\sf{(10+x)\:m}$

∠BAD = 45°

So,

$\sf{Tan45^\circ = \dfrac{BD}{AB}}$

= $\sf{1 = \dfrac{10+x}{10\sqrt{3}}}$

= $\sf{10\sqrt{3} = 10+x}$

= $\sf{10\sqrt{3} - 10 = x}$

= $\sf{10(\sqrt{3} - 1) = x}$

=> $\sf{x = 10(\sqrt{3} - 1)}$

=> $\sf{x = 10(1.73 - 1)}$

=> $\sf{x = 10\times 0.73}$

=> $\sf{x = 7.3\:m}$

Therefore the height of the tower is 10(√3 - 1) m or 7.3 m.

______________________________________

Important!!!

Trigonometric table:-

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} \bf Angles & \bf 0^{o} & \bf 30^{o} & \bf 45^{o} & \bf 60^{o} & \bf 90^{o} \\\cline{1-6} \tt Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}}& \dfrac{\sqrt{3}}{2}& 1\\\cline{1-6} \tt cos \theta & 1 & \dfrac{\sqrt{3}}{2} &\dfrac{1}{\sqrt{2}}&\dfrac{1}{2}&0\\\cline{1-6} \tt tan \theta & 0 & \dfrac{1}{\sqrt{3}} & 1& \sqrt{3} & \infty \\\cline{1-6} \tt cosec \theta & \infty & 2 & \sqrt{2} & \dfrac{2}{\sqrt{3}} &1\\\cline{1-6} \tt sec \theta & 1 & \dfrac{2}{\sqrt{3}} & \sqrt{2} & 2 & \infty \\\cline{1-6} \tt cot \theta & \infty & \sqrt{3} &1& \dfrac{1}{\sqrt{3}} &0\\\cline{1-6}\end{array}$

______________________________________