Question

1. Point B on the ground is the base of a vertical pillar AB. Two points D and C on the ground lie on the straight line BCD. From one of the two points C and D, the elevation angle to point A is 45º and from the other it is 53°. If CD= 3 m, find the height of the pillar AB​

Answer 1

Given:

B is the base of the pillar AB

BCD is a straight line

The angle of elevation to point A is 45º and 53° from points C and D.

CD = 3 m

To find:

The height of the pillar AB

Solution:

Formula to be used for solving the problem:

• Trigonometric ratio of right-angled triangles: tan θ = $\frac{Perpendicular}{Base}$

Now,

Referring to the figure attached below, we will consider,

AB = the height of the pillar = "h"

BC = "x"

∠ACB = 53° = angle of elevation from point C to point A

∠ADB = 45° = angle of elevation from point D to point A

Considering ΔABC and applying the formula, we get

tan 53° = $\frac{AB}{BC}$

⇒ tan 53° = $\frac{h}{x}$

⇒ 1.327 = $\frac{h}{x}$

⇒ h = 1.327x ....... (i)

Considering ΔDB and applying the formula, we get

tan 45° = $\frac{AB}{BD}$

⇒ tan 45° = $\frac{AB}{BC\: +\: CD}$

⇒ tan 45° = $\frac{h}{x\:+\:3}$

⇒ 1 = $\frac{h}{x\:+\:3}$

⇒ x + 3 = h

we will substitute the value of h from eq. (i)

⇒ x + 3 = 1.327x

⇒ 1.327x - x = 3

⇒ 0.327x = 3

⇒ x = $\frac{3}{0.327}$

⇒ x = 9.17 m

We will substitute the value of x = 9.17 m in eq. (i)

h = 1.327x = 1.327 × 9.17 = 12.16 m

Thus, the height of the pillar AB is 12.16 m.

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