A horizontal park is in the shape of a triangle OAB with $\mathrm{AB}=16$. A vertical lamp post OP is erected at the point O such that $\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ}$ and $\angle \mathrm{PCO}=45^{\circ}$, where C is the midpoint of AB. Then $(\mathrm{OP})^{2}$ is equal to

(B) $\frac{32}{\sqrt{3}}(2-\sqrt{3})$

(A) $\frac{32}{\sqrt{3}}(\sqrt{3}-1)$

(C) $\frac{16}{\sqrt{3}}(\sqrt{3}-1)$

(D) $\frac{16}{\sqrt{3}}(2-\sqrt{3})$

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Heights and Distances

A horizontal park is in the shape of a triangle OAB with

AB = 16

. A vertical lamp post OP is erected at the point O such that

∠ PAO = ∠ PBO = 1 5^{\circ}

and

∠ PCO = 4 5^{\circ}

, where C is the midpoint of AB. Then

(OP)^{2}

is equal to

Answer: (B)

Let

h = OP

. From the right triangle

OPC

with

∠ PCO = 4 5^{\circ}

, we have

OC = h

. From the right triangle

OPA

with

∠ PAO = 1 5^{\circ}

, we have

OA = h cot 1 5^{\circ} = h (2 + 3)

. Since C is the midpoint of AB and

AB = 16

, we have

AC = 8

. In triangle

OAC

, using the Pythagorean theorem:

OA^{2} = OC^{2} + AC^{2}

, so

h^{2} (2 + 3)^{2} = h^{2} + 64

. Simplifying:

h^{2} [(2 + 3)^{2} - 1] = 64

, which gives

h^{2} [7 + 43 - 1] = 64

, so

h^{2} (6 + 43) = 64

. Therefore,

h^{2} = \frac{64}{6 + 4 3} = \frac{32}{3 + 2 3} = \frac{32 ( 3 - 2 3 )}{( 3 + 2 3 ) ( 3 - 2 3 )} = \frac{32 ( 3 - 2 3 )}{9 - 12} = \frac{32 ( 2 3 - 3 )}{3} = \frac{32}{3} (2 - 3)

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option