Question

from a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the of a 20 m high buildingare 45° and 60° respectively.Find the hight of the tower​

Answer 1

Answer:

Let DC be the tower and BC be the building. Then,

∠CAB=45

o

,∠DAB=60

o

,BC=20 m

Let height of the tower, DC=h m.

In right △ABC,

tan45

o

=

AB

BC

1=

AB

20

AB=20 m

In right △ABD,

tan60

o

=

AB

BD

3

=

20

h+20

h=20(

3

−1) m

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

A transmission tower, stands on the top of a 20 m high building. From a point on the ground, the angle of elevation of the top of the transmission tower is 60° and from the same point the angle of elevation of the bottom of the transmission tower is 45°.

Let assume that AB represents the height of high building and BC represents the transmission tower.

So, AB = 20 m

Let assume that the height of transmission tower be h m.

So, BC = h m.

Let D be any point on the ground such that the angle of elevation of the top of the transmission tower is 60° and from the same point the angle of elevation of the bottom of the transmission tower is 45°.

So, it means ∠ADB = 45° and ∠ADC = 60°.

Further, assume that D point is x m away from the foot of building AB.

So, we have with us the following data :

↝ AB = 20 m

↝ BC = h m

↝ AD = x m

↝ ∠ADB = 45°

↝ ∠ADC = 60°

Now, In right angle triangle ADB

$\rm :\longmapsto\:tan45 \degree \: = \: \dfrac{AB}{AD}$

$\rm :\longmapsto\:1 = \dfrac{20}{x}$

$\bf\implies \:\boxed{ \tt{ \: \: x \: = \: 20 \: m \: \: }} - - - (1)$

Now, In right-angle triangle ACD

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{AC}{AD}$

$\rm :\longmapsto\: \sqrt{3} \degree \: = \: \dfrac{h + 20}{x}$

$\rm :\longmapsto\: \sqrt{3} \degree \: = \: \dfrac{h + 20}{20}$

$\rm :\longmapsto\:20 \sqrt{3} = 20 + h$

$\rm :\longmapsto\:20 \sqrt{3} - 20 = h$

$\bf :\longmapsto\:h \: = \: 20 (\sqrt{3} - 1) \: m$

OR

$\bf :\longmapsto\:h \: = \: 20 (1.732 - 1) \:$

$\bf :\longmapsto\:h \: = \: 20 (0.732) \:$

$\bf\implies \:h = 14.64 \: m$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: h = 20( \sqrt{3} - 1) \: m \: \: or \: \: 14.64 \: m \: }}}$

More to know :-

$\blue{\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}$