Question

There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are the points directly opposite to each other on the two banks, and in a line with the tree. If the angles of elevation of the top of tree from P and Q are 30° and 45° respectively, find the height of the tree.

Answer 1

$\huge \frak \pink{Solution}$

Let OA be the tree of height (h) metre.

In Triangle POA and QOA, we have

$\small\leadsto \sf\tan30° = \frac{OA}{OP} \: and \: \tan45° = \frac{OA}{OQ}$

$\implies \sf \frac{1}{ \sqrt{3} } = \frac{h}{OP} \: and \: 1 = \frac{h}{OQ}$

$\implies \sf OP= \sqrt{3} h \: and \: oq = h$

$\implies \sf OP+OQ = \sqrt{3} h + h$

$\implies \sf PQ= ( \sqrt{3} + 1)h$

$\implies \sf 100 = ( \sqrt{3} + 1)h \: \: \\ \sf [∵PQ = 100m ]$

$\implies \sf h = \frac{100}{ \sqrt{3} + 1 } m$

$\implies \sf h = \frac{100( \sqrt{3} + 1) }{2} m$

$\small\implies \sf h = 50(1.732 - 1)m = 36.6m$

Hence, the height of the tree is 36.6 m.

Answer 2

${ \large{ \sf{ \underbrace{\underline{\bigstar \: Answer}}}}}$

$\sf\color{red}{Let\: PQ \:be \:the\: width \:of \:the \:river \:and \:RS\: be \:the\: height\: of \:the \:tree\: on\: the\: island}$

$\boxed{\bf{In\: rt.\:△\:PRS}}$

$\sf{x = RS cot 30°}$

$\sf{x = RS √3}$

$\sf{x = √3 RS .(i)}$

$\boxed{\bf{In \:rt.\:△\:RSQ,}}$

$\sf{SQ = RS cot 45°}$

$\sf{(100-x) = RS}$

$\sf{x=100 - RS .(ii)}$

$\boxed{\bf{Equating\: (i) \:and\: (ii) \:we\: have:}}$

$\sf{√3 RS = 100 - RS}$

$\sf{2.73 RS = 100}$

$\sf{RS = 36.63 m}$

$★ \pmb {\red{Therefore\: the\: height\: of\: the \:tree \:is \:36.33 m. }}★$

Thankyou :)

~I wanted to make some changes in last answer on your question but someone reported so can't sorry :(