Question

A telegraph post is bent at a point above the ground due to storm. Its top just meets the ground at a
distance of 8 root 3 metres from its foot and makes an angle of 30 degree. Calculate at what height the post is bent
and what is the height of the post.

Answer 1

AnswEr :

• Height of Post Bent is 16 m.

• Height of the Post is 24 m.

Diagram :

$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,3.9){\large\sf{D}}\put(7.7,2.9){\large\sf{B}}\put(7.7,1){\large\sf{C}}\put(10.6,1){\large\sf{A}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){3}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.4,0.7){\sf{\large{8 \sqrt{3} }}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{30^{\circ} }}\end{picture}$

Here,

• AB = Bent part

• AC = 8√3 m

Let us consider that the height of the Post is AB + BC.

In ∆ABC,

$\longrightarrow\sf \: \tan ({30}^{ \circ}) = \dfrac{h}{ 8\sqrt{3} }$

$\longrightarrow\sf \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{ 8\sqrt{3} }$

$\longrightarrow\sf \: h = \dfrac{ 8\sqrt{3} }{ \sqrt{3} }$

$\longrightarrow \large\sf \: \blue{h = 8 \: m}$

Now,

Apply Pythagoras theorem,

$\longrightarrow \sf \: {AB}^{2} = {BC}^{2} + {AC}^{2}$

$\longrightarrow \sf \: {AB}^{2} = {(8)}^{2} + { (8\sqrt{3} )}^{2}$

$\longrightarrow \sf \: {AB}^{2} = 64 + 192$

$\longrightarrow \sf \: {AB}^{2} = 256$

$\longrightarrow \sf \large \: \purple{{AB} = 16 \: m}$

We know that,

$\longrightarrow$ Height of the Post = AB + BC

$\longrightarrow$ Height of the Post = 16 + 8

$\therefore$ Height of the Post = 24 m

• Height of Post bent = AB = 16 m

Hence solved!