Question

A man observes the angle of elevation of the top of a building to be 30.he walks towards it ina horixontal line through its base. on covering 60 m.the amgle of elevation changea to 60 find the height of the building correct to the mearest metre

Answer 1

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Answer 2

Answer:

Height of Building = 51.96 m

Step-by-step explanation:

Given: Angle of elevation of building from point A is 30°

           Angle of elevation from point P is 60°

           Distance AP = 60 m

To find: Height of Building

Figure is attached

let height, CB = h and distance DB = x

In Δ BDC

using trigonometric ratio we get,

$tan\,60^{\circ}\,=\,\frac{h}{x}$

$\sqrt{3}\,=\,\frac{h}{x}$

$x\,=\,\frac{h}{\sqrt{3}}$ ......... (1)

In Δ BAC

again using trigonometric ratio we get,

$tan\,30^{\circ}\,=\,\frac{h}{60+x}$

$\frac{1}{\sqrt{3}}\,=\,\frac{h}{60+x}$

$\frac{1}{\sqrt{3}}\times(60+x)\,=\,h$

Now put value of x from equation (1)

$\frac{1}{\sqrt{3}}\times(60+\frac{h}{\sqrt{3}})\,=\,h$

$\frac{60}{\sqrt{3}}+\frac{h}{3}\,=\,h$

$h-\frac{h}{3}=20\sqrt{3}$

$\frac{2h}{3}=20\sqrt{3}$

h = 30√3 m

h = 51.96 m

Therefore, Height of Building = 51.96 m