Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from $30^{\circ} t o 45^{\circ}$ .

Open in App

Solution

Let AB be the building of height h m.

Let the two points be C and D such that CD = 40 m, ADB = 30^o and ACB = 45^o

$i n space increment A B C comma space space space space space space space space fraction numerator A B over denominator B C end fraction equals tan space 45 degree equals 1 rightwards double arrow space space space space space space B C equals h i n space increment A B D space space space space space space space space space fraction numerator A B over denominator B D end fraction equals tan space 30 degree rightwards double arrow space space space space fraction numerator h over denominator 40 plus h end fraction equals fraction numerator 1 over denominator square root of 3 end fraction rightwards double arrow space space space square root of 3 h equals 40 plus h space space therefore space space space space h equals fraction numerator 40 over denominator square root of 3 minus 1 end fraction equals fraction numerator 40 over denominator 0.732 end fraction equals 54.64 space m$

Hence, height of the building is 54.64 m.

Suggest Corrections

Similar questions

Q. Find the height of the chimney when it is found that on walking towards it

50

m in the horizontal line through its base, the angle of elevation of its top changes from

30^{\circ}

and to

60^{\circ}

Q. Find the height of a tree when it is found that on walking away from it 20 m. in a horizontal line through its base, the elevation of its top changes from

60^{0}

30^{0}

Q. It is found that on walking x meters towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60°. The height of the chimney is

(a)

3 \sqrt{2} x

(b)

2 \sqrt{3} x

(c)

\frac{\sqrt{3}}{2} x

(d)

\frac{2}{\sqrt{3}} x

A man observes the angle of elevation of the top of a building to be $30^{\circ}$ . He walks towards it in horizontal line through its base. On covering 60 m the angle of elevation changes to $60^{\circ}$ . Find the height of the building correct to the nearest metre.

Q.
The angle of elevation of the top of a hill when observed from a certain point on the horizontal plane through its base is

30^{0}

. After walking 120 meters towards it on level ground the elevation is found to be

60^{0}

. Find the height of the hill(in meters).

Join BYJU'S Learning Program

Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30∘ to 45∘.

Let AB be the building of height h m. Let the two points be C and D such that CD = 40 m, ADB = 30o and ACB = 45o Hence, height of the building is 54.64 m.

Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from $30^{\circ} t o 45^{\circ}$ .