Question

① The
The angle of elevation of the top
of à tower from a point on the
bound which is 30m away from
foot
the foot of tower is 30° find the height
​

Answer 1

Answer:

60 °

Step-by-step explanation:

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Answer 2

Given :-

• Angle ACB = 30°
• BC = 30 m

To find :-

• Height of tower = AB = ?

Solution :-

In the attached figure, ∆ABC is a right angled triangle.

So, if we find out tan θ then we can find the value of AB.

$\red{ \sf{tan \theta} = \dfrac{opposite \: side}{adjacent \: side} }$

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

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

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

$\Rightarrow \sf{AB = \dfrac{30}{ \sqrt{3} } }$

On rationalising the Denominator,

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

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

$\Rightarrow \: \sf{AB = 10 \sqrt{3} \: m}$

Since √3 = 1.73,

∴ The height of the tower

= 10√3 m

= 17.3 m

Know more:

1. Sin theta can be written as Opposite side divided by Hypotenuse.
2. Cos theta can be written as Adjacent side divided by Hypotenuse.

Where theta is an angle in the right angled triangle.

This question can also be solved by taking cot theta and take reciprocal on RHS instead of tan theta.