Question

ABCD is a rectangle field. A vertical lamp post of height 12 m stands at the corner A. If the angle of elevation of its top from B is 60° and from C is 45°, then the area of the field is

(a) 48\(\sqrt2\) sq m

(b)  48\(\sqrt3\) sq m

(c)  12\(\sqrt2\) sq m

(d)  12\(\sqrt3\) sq m

Answer 1

(a) 48\(\sqrt2\) sq m

Let ABCD be the rectangular field and EA be the lamp post that stands at corner A.

Since AE is a vertical pole, it is perpendicular to all lines in the plane of the rectangular field, i.e., EA is perp. to AB, BC, CD, DA.

∴ ∠EAD = 90°.

Given,

EA = 12 m

Join EB, EC, and AC.

Given,

∠EBA = 60°, ∠ACE = 45°.

In rt. Δ ABE,

tan 60° = \(\frac{AE}{AB}=\frac{12}{AB}\)

⇒ \(\sqrt3=\frac{12}{AB}\)

⇒ AB = \(\frac{12}{\sqrt3}\) = 4\(\sqrt3\) m.

In rt. Δ ACE,

tan 45° = \(\frac{AE}{AC}\)

⇒ \(\frac{12}{AC}=1\) 

⇒ AC = 12 m.

In Δ ABC,

BC = \(\sqrt{AC^2-AB^2}\)   (Pythagoras’ Th.)

\(= \sqrt{144\,-\,48}\)

= \( \sqrt{96}\) m

= 4\(\sqrt6\) m

∴ Area of the rectangular field = AB × BC

= \(4\sqrt3\,\times\,4\sqrt6\)

= \(48\sqrt2\)  sq. m.