Question

$\bold{\mathfrak{Hey \ Mates}}$

Here's a Maths question for uh guys..!

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 30°​

Answer 1

Given :-

• AC = 20 m
• $\angle$C = 30°

To find :-

• The height of the pole = ?

Figure :-

$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.7,2){\sf{\large{?}}}\put(9.3,2){\sf{\large{20 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{30^{\circ} }}\end{picture}$

Solution :-

In ∆ ABC,

➯ $\sf \dfrac{AB}{BC} = sin \:30^{\circ}$

➯ $\sf \dfrac{AB}{20} = \dfrac{1}{2}$

➯ $\sf AB =\dfrac{20}{2}$

➯ $\underline{\boxed{\red{ \textbf{AB = 10\:m}}}}$ $\dagger$

Therefore the height of the pole is 10 m.

$\rule{250}3$

Extra brainly knowledge :-

$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$

$\rule{250}3$

Answer 2

Answer:

The height of the pole : 10 m

Given:

➞A circus artist is climbing a 20 m long rope.

➞The angle made by the rope with the ground level is 30°.

To Find:

The height of the pole.

Solution:

We are given ,

➞A circus artist is climbing a 20 m long rope.

➞The angle made by the rope with the ground level is 30°.

( Refer the attachment ).

Let PQ be the vertical pole and RP be the rope.

Then,

∠PRQ = 30°. ( Given )

and

PR = 20 m. ( Given )

We know,

${\bold{\green{\boxed{\pink{\star{\rm{\sin(\theta) = \dfrac{Opposite \ side }{hypothenuse}}}}}}}}$

Now,

In right angled triangle PRQ,

${\implies{\rm{ \sin(30 \degree) = \dfrac{PQ}{PC}}}}$

but,

${\bold{\green{\boxed{\pink{\star{\rm{\sin(30 \degree) = \dfrac{ 1}{2}}}}}}}}$

${\implies{\rm{\dfrac{ 1}{2} = \dfrac{PQ}{20}}}}$

PQ = 10.

Therefore, the height of the pole is 10 m.