Question

Answer plzzz❤️❤️
will mark as BRAINLIST ✌️✌️​

Answer 1

Answer:

hope it helps........

Answer 2

Let the boy be at a distance d fron the temple initially. Then,

$30-1.5=d\tan30^{\circ}\\\\\\d=28.5\sqrt{3}$

After moving x distance towards the temple,

$d-x=\dfrac {28.5}{\tan60^{\circ}}\\\\\\28.5\sqrt{3}-x=\dfrac {28.5}{\sqrt3}\\\\\\x=28.5\left (\sqrt3-\dfrac {1}{\sqrt3}\right)\\\\\\\underline {\underline {x=32.9\ m}}$