Question

the angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower, is 30°.Find the height of the tower.​

Answer 1

Answer:

Height of the tower is 10√3 m or 17.32 m.

$\rule{100}2$

Step-by-step explanation:

Consider the height of tower be AB. The angle of elevation of the top of the tower from a point (consider that point be C) on the ground, which is 30 m away. Means, BC = 30 m.

Angle of elevation from the top of a tower to a point on the ground is 30°.

We have to find the height of the tower.

[ Refer the attachment for figure ]

In ∆ABC

$\implies\:\sf{\dfrac{P}{B}\:=\:tan30^{\circ}}$

$\implies\:\sf{\dfrac{AB}{BC}\:=\:tan30^{\circ}}$

We know that -

BC = 30 m

tan30° = 1/√3

Now, put the known values above formula.

$\implies\:\sf{\dfrac{AB}{30}\:=\:\dfrac{1}{\sqrt{3}}}$

Cross multiply them

$\implies\:\sf{AB\sqrt{3}\:=1(30)}$

$\implies\:\sf{AB\sqrt{3}\:=\:30}$

We have to find AB. So, take √3 on right hand side.

$\implies\:\sf{AB\:=\:\dfrac{30}{\sqrt{3}}}$

Rationalize it

$\implies\:\sf{AB\:=\:\dfrac{30}{\sqrt{3}}\:\times\:\dfrac{\sqrt{3}}{\sqrt{3}}}$

We know that √3 × √3 = (√3)² = 3

$\implies\:\sf{AB\:=\:\dfrac{30\sqrt{3}}{3}}$

$\implies\:\sf{AB\:=\:10\sqrt{3}}$

Value of √3 = 1.732

$\implies\:\sf{AB\:=\:10(1.732)}$

$\implies\:\sf{AB\:=\:17.320}$

$\implies\:\sf{AB\:=\:17.32\:m}$

Answer 2

$\Large\underline{\underline{\sf{Given}:}}$

• Angle of elevation of the top of tower and point on ground is 30° .
• distance of foot of tower and point is 30m .

$\large\boxed{\underline{\mathcal{\red{Q}\green{u}\pink{e}\orange{s}\blue{ti}\red{on.??}}}}$

• we have to Find Height of the tower ?

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• Tan@ = Perpendicular/Base
• Tan30° = 1/√3
• √3 = 1.732

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$\Large\bold\star\underline{\underline\textbf{Solution(1)}}$

$\red{\textbf{Refer To image First}}$

we have ,

→ angle ACB = 30°

→ BC = 30m

→ AB = Height of tower = h metre (Let) .

As we know that,

$\tan( \alpha) = \frac{p}{b} \\ \\ so \\ \\ \red{\boxed\implies} \: \tan(30) = \frac{AB}{BC} \\ \\ \red{\boxed\implies} \: \frac{1}{ \sqrt{3} } = \frac{h}{30} \\ \\ \red{\boxed\implies} \: h = \frac{30}{ \sqrt{3} } \\ \\ \green{\textbf{Rationalize the Denominator}} \\ \\ \red{\boxed\implies} \: h = \frac{30 \sqrt{3} }{3} \\ \\ \red{\boxed\implies} \: h = 10 \sqrt{3} \\ \\ \red{\boxed\implies} \: h = 10 \times 1.732 \\ \\ \large\boxed{\bold{h = 17.32m}}$

Hence, Height of Tower will be 17.32 m..

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$\Large\bold\star\underline{\underline\textbf{Solution(2)}}$

Lets Try it with Right angle ∆ ratio Method now ,

As we know that, Angle of ∆ are in ratio = 1 : 2 : 3

so,

Their sides are in ratio of = 1: √3 : 2

so, we have ,

$\pink{\large\boxed{\bold{30:60:90 = 1 : \sqrt{3} : 2}}}$

it is given that ,

$\red\leadsto \: \sqrt{3} = 30m$

we have To Find 1 unit, that is opposite of angle 30° . as you can see in diagram .

so,

$\red\leadsto \: \sqrt{3} = 30m \\ \\ \red\leadsto \: 1 = \frac{30}{ \sqrt{3} } \\ \\ rationalizing \\ \\ \red\leadsto \: 1 = \frac{30 \sqrt{3} }{3} \\ \\ \red\leadsto1 = 10 \sqrt{3} \\ \\ \red\leadsto1 = 10 \times 1.732 \\ \\ \red\leadsto \: \orange{\large\boxed{\bold{1 = 17.32m}}}$

Hence, Height of Tower that is opposite of 30° is 1.732m.

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$\large\underline\textbf{Hope it Helps You.}$