Question

27. From a point on the ground, the angles of elevation of the top and bottom of
a transmission tower fixed at the top of a 20m high building are 60% and 45°
respectively. Find the height of the transmission tower
A
0=60
B
20m
09
Ist
1 Science E SI OP
C С
D​

Answer 1

Correct Statement :-

• From a point on the ground, the angles of elevation of the top and bottom of a transmission tower fixed at the top of a 20m high building are 60° and 45° respectively. Find the height of the transmission tower.

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

Let BC be the height of building.

• So, BC = 20 m

Let Height of transmission tower be CD = 'x'.

Let A be the point on the ground at a distance of 'y' meter from the foot of the building B.

The angle of elevation from point A to C and D be 45° and 60°.

So,

$\rm :\longmapsto\:In \: \triangle \: ABC$

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

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

$\bf\implies \:y = 20 \: m - - - (1)$

Now,

$\rm :\longmapsto\:In \: \triangle \: ADB$

$\rm :\longmapsto\:tan60 \degree \: = \dfrac{BD}{AB}$

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

$\rm :\longmapsto\: \sqrt{3} = \dfrac{x + 20}{20} \: \: \: \: \: \: \: \: \{using \: (1) \}$

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

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

$\bf\implies\:x=20(\sqrt{3}-1)=20(1.732 -1)=14.64\:m$

$\boxed{\bf\: Hence \: height \: of \: transmission \: tower = 14.64 \: m}$

Additional Information :-

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