Height and Distance

Let AB be the tower and CD be the electric pole.
Then, $$\angle ACB = {60^ \circ },$$     $$\angle EDB = {30^ \circ }$$     and AB = 15 m
Let CD = h
Then, BE = (AB - AE) = (AB - CD) = (15 - h)
$$\frac{{AB}}{{AC}} = \tan {60^ \circ } = \sqrt 3 $$
$$ ⇒ AC = \frac{{AB}}{{\sqrt 3 }} = \frac{{15}}{{\sqrt 3 }}$$
$${\text{And, }}\frac{{BE}}{{DE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ DE = \left( {BE \times \sqrt 3 } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {15 - h} \right)$$
$${\text{So, }}AC = DE$$
$$ ⇒ \frac{{15}}{{\sqrt 3 }} = \sqrt 3 \left( {15 - h} \right)$$
$$ ⇒ 3h = \left( {45 - 15} \right)$$
$$ ⇒ h = 10{\text{ m}}$$

$${\text{Given tan }}{x^ \circ } = \frac{2}{5}$$     and AF = 200 meter
$$\therefore \frac{2}{5} = \frac{{TF}}{{AF}}$$
$$ ⇒ TF = \frac{{2 \times 200}}{5}$$
$$ ⇒ TF = 80{\text{ m}}$$
$${\text{We have, BF = 80 m}}$$
$$\therefore \tan {\text{ }}{y^ \circ } = \frac{{TF}}{{BF}}$$
$$ ⇒ \tan {\text{ }}{y^ \circ } = \frac{{80}}{{80}}$$
$$ ⇒ \tan {\text{ }}{y^ \circ } = 1 = \tan {45^ \circ }$$
$$\therefore {y^ \circ } = {45^ \circ }$$

$${\text{Let, }}OP = {\text{ }}a$$
$${\text{tan }}{60^ \circ } = \frac{H}{a}$$
$$ ⇒ H = \sqrt 3 a$$
$$ ⇒ \frac{H}{{\sqrt 3 }} = a.....(i)$$
$$tan{45^ \circ } = 1$$   $$ = \frac{H}{{a + 100\left( {3 - \sqrt 3 } \right)}}$$
$$ ⇒ a + 100\left( {3 - \sqrt 3 } \right) = H$$
From (i) $$\frac{H}{{\sqrt 3 }} + $$   $$100\left( {3 - \sqrt 3 } \right)$$   = H
$$ ⇒ H + 300\sqrt 3 - 300 = \sqrt 3 H$$
$$ ⇒ 300\sqrt 3 - 300 = \sqrt 3 H - H$$
$$ ⇒ \left( {\sqrt 3 - 1} \right)H = 300\left( {\sqrt 3 - 1} \right)$$
$$ ⇒ H = 300{\text{ m}}$$

Let AB be lighthouse and P and Q are two ships on it's opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB =×and QB = y

$${\text{Now in right }}\Delta APB$$
$$\tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}}$$
$$ ⇒ \tan {45^ \circ } = \frac{h}{x} ⇒ 1 = \frac{h}{x}$$
$$ ⇒ x = h\,............(i)$$
$${\text{Similarly in right }}\Delta AQB,$$
$$\tan {30^ \circ } = \frac{{AP}}{{QB}} = \frac{h}{y}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{h}{y}$$
$$ ⇒ y = \sqrt 3 \,h\,..............(ii)$$
$${\text{Adding (i) and (ii)}}$$
$$\therefore PQ = x + y$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = h + \sqrt 3 \,h$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt 3 + 1} \right)\,h$$

Let AB is girls and CD is lamp-post AB = 1.5
which casts her shadow EB

$$\therefore EB = 4.5\,m,\,BD = 3\,m$$
$${\text{Now in }}\Delta AEB$$
$$\tan \theta = \frac{{AB}}{{BE}} = \frac{{1.5}}{{4.5}} = \frac{1}{3}$$
$${\text{and in }}\Delta CED$$
$$\tan \theta = \frac{{CD}}{{ED}} ⇒ \frac{1}{3} = \frac{h}{{4.5 + 3}}$$
$$ ⇒ \frac{1}{3} = \frac{h}{{7.5}}$$
$$ ⇒ h = \frac{{7.5}}{3} - 2.5\,m$$
$$\therefore {\text{Height of lamp - post}} = {\text{2}}{\text{.5}}\,m$$

$${\text{Original tree height}} = {\text{h}} = {\text{MQ}}$$
$${\text{New}}\,{\text{tree}}\,{\text{height}} = PQ$$
$${\text{in}}\,\Delta MQN,\,\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{NQ}}$$
$$\therefore MQ = \frac{{300}}{{\sqrt 3 }}$$
$${\text{in}}\,\Delta PQN,\,\tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{NQ}}$$
$$\therefore PQ = 300\sqrt 3 $$
$${\text{Tree}}\,{\text{grew}} = PQ - MQ$$
$$\therefore {\text{Tree}}\,{\text{grew}}$$
$$ = 300\sqrt 3 - \frac{{300}}{{\sqrt 3 }}$$
$$ = 300\frac{2}{{\sqrt 3 }}$$
$$ = 3 \times 100 \times \frac{2}{{\sqrt 3 }}$$
$$ = 200\sqrt 3 \,{\text{m}}$$

Let PQ be three and M and N be positions where Mphan stands.
$${\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{MQ}}$$
$$\therefore PQ = MQ$$
$$\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{NQ}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{10 + MQ}}$$
$$\therefore 10 + MQ = \sqrt 3 \,PQ$$
$$\therefore 10 + PQ = \sqrt 3 \,PQ$$     (As, PQ = MQ)
$$\therefore PQ = \frac{{10}}{{\sqrt 3 - 1}}\,{\text{cm}}$$

$${\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{NQ}}$$
$$\therefore {\text{Tree}}\,{\text{height}}$$
$$ = PQ = NQ$$
$$ = \left( {240 + MQ} \right)$$
$$\tan {60^ \circ } = \sqrt 3 $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{MQ}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 + MQ}}{{MQ}}$$
$$\therefore 240 + MQ = \sqrt 3 \,MQ$$
$$\therefore MQ = \frac{{240}}{{\sqrt 3 - 1}}\,{\text{ft}}$$
$$\therefore NQ = 240 + MQ$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240\sqrt 3 }}{{\sqrt 3 - 1}}\,{\text{ft}}$$
= Rohit was this much far away initially

Let the tree break at height h cm from ground at point M.
The broken part makes angle of 30°
∴ Broken Part MP = MN = 24 - h
$${\text{in}}\,\Delta MCQ,$$
$$\sin {30^ \circ } = \frac{1}{2} = \frac{{MQ}}{{MN}} = \frac{{\text{h}}}{{24 - {\text{h}}}}$$
$$\therefore 24 - {\text{h}} = 2{\text{h}}$$
∴ h = 8 cm = Tree breaks at this height

In ∠ABR, ∠ARB = 90° and ∠BAR = 45°
Sum of angles of a triangle = 180°
So ∠ABR = 180 - 90 - 45 = 45°
∴ BR = AR
AS = RQ = 10m
Also, BR = BQ - RQ = 25 - 10 = 15m
∴ AR = 15m = Distance between houses