Question

From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of observation.

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Answer 1

To find → Height of tower (AB)

Let AB =h m

In △ABC, by Trigonometry

tan60∘=BCAB

3=BCh⇒BC=(3h)m

Since BCDE is a ||gm, so BC=DE and CD = BE

∴DE=(3h)m;BE=40m

Now in △ADE

tan30∘=DEAE

31=3hh−40

h=3h−120

2h=120m

h=60m

Height of tower = 60 m.

To find → Horizontal Distance from point of observation (BC)

BC=3h=360=203m

BC=203m=34.60m

Step-by-step explanation:

I hope this will help you

Answer 2

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

Let assume that CD be the height of the tower.

Now, From a point A, on the ground, the angle of elevation of the top of a tower is observed to be 60°.

Now, From a point B, 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 30°.

Let assume that

Height of tower, CD = h m

AC = BE = x m.

Now, In right-angled triangle ACD

$\rm \: tan60 \degree \: = \: \dfrac{CD}{AC}$

$\rm \: \sqrt{3} \: = \: \dfrac{h}{x}$

$\rm\implies \:h = \sqrt{3}x - - - (1)$

Now, In right-angled triangle BED

$\rm \: tan30 \degree \: = \: \dfrac{ED}{BE}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{h - 40}{x}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3}x - 40}{x}$

$\rm \: 3x - 40 \sqrt{3} = x$

$\rm \: 3x - x= 40 \sqrt{3}$

$\rm \: 2x= 40 \sqrt{3}$

$\rm\implies \: x= 20 \sqrt{3}$

On substituting the value of x in equation (1), we get

$\rm \: h = 20 \sqrt{3} \times \sqrt{3}$

$\rm\implies \:h \: = \: 60 \:$

Hence,

The height of the tower = 60 m

and

Its horizontal distance from the point of observation is 20√3 m or 20 × 1.73 = 34.6 m

$\rule{190pt}{2pt}$

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