Question

the angle of depression of a vehicle on the ground from the top of a tower is 60.if a vehicle is at a distance of 100 m away from the building,find the height of the tower.​

Answer 1

Explanation:

Given,

The angle of depression, ∠ SPR = 60°

So,

The angle of elevation, ∠ QRP = Angle of depression, ∠ SPR = 60°.

Now, in right triangle PQR,

QR = 100m

∠ R = 60°

PQ = h (in m)

Let ∠ R = θ = 60°

We know,

tan θ = (Opposite Side/Adjacent Side)

Tan 60 = PQ/RQ = h/100

√3 = h/100

Or

h = √3 x 100

h = 173.20 m

Hence, the height of the tower is 173.20 meters.

Answer 2

Given :

• Angle = 60°
• Distance of the car from the builiding = 100 m

To Find :

• The height of the tower

Solution :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){4}}\qbezier(0,0)(0,0)(4,4)\put(-0.7,-0.2){\bf C}\put(4.3,-0.2){\bf B}\put(4.2,4.2){\bf A}\qbezier(0.7,0)(0.8,0.35)(0.5,0.5)\put(0.9,0.3){ \bf 60^{\circ} }\end{picture}$

Let

• AB be the height of the tower
• BC be the distance of car from the tower

Tanθ is given by ,

$\\ \star \: \boxed{\purple{\sf{ \tan \theta = \frac{opposite \: side}{adjacent \: side} }}}$

In this case ,

• θ = 60°
• BC becomes the adjacent side
• AB becomes the opposite side

Substituting the values we have ,

$\\ : \implies \sf \: \tan( {60}^{ \circ} ) = \frac{AB}{100}$

$\\ : \implies \sf \: \sqrt{3}= \frac{AB}{100}$

$\\ :\implies{\boxed{\sf{\pink{AB = 100\sqrt{3}\: m}}}}\: \bigstar$

$\\ \therefore{\underline{\sf{Hence , \: The \: height \: of \: the \: tower \: is \: \bold{100\sqrt{3} \: m}}}}$