Height and Distance

$${\text{AC = 30 meter}}$$
$${\text{AB = 15 meter}}$$
$$\angle {\text{ACB}} = \theta $$
$$\therefore \sin \theta = \frac{{AB}}{{AC}} = \frac{{15}}{{30}} = \frac{1}{2}$$
$$ ⇒ \sin \theta = \sin {30^ \circ }$$
$$ ⇒ \theta = {30^ \circ }$$

$${\text{tan }}\theta = \frac{H}{{{m^2}}}$$
$$ ⇒ \tan \left( {90^ \circ - \theta } \right) = \frac{H}{{{n^2}}}$$
$$ ⇒ \cot \theta = \frac{H}{{{n^2}}}$$
$$ ⇒ \tan \theta .\cot \theta = \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}}$$
$$ ⇒ \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} = 1$$
$$ ⇒ {H^2} = {m^2}{n^2} $$
$$ ⇒ H = mn$$

$${\text{Here, AD is parallel to BC and AC is a transversal}}{\text{.}}$$
$$ ⇒ {\theta _1} = {\theta _2}$$

$${\text{Let, }}AC = H{\text{ }}m$$
$$OA = {\text{ }}a{\text{ }}m$$
$$\tan {\text{ }}{30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{H}{a}$$
$$\tan {60^ \circ } = \sqrt 3 $$    $$ = \frac{{H + 200 + 200}}{a}$$
$$ ⇒ \frac{{H + 400}}{{\sqrt 3 H}} = \sqrt 3 $$
$$ ⇒ H + 400 = 3H$$
$$ ⇒ 2H = 400\,m$$
$$ ⇒ H = 200\,m$$
Height of cloud above lake = 200 m

Let AB be the tower and C and D be the two positions of the boats.
Then, $$\angle {\text{ACB = }}{45^ \circ },$$    $$\angle ADB = {30^ \circ }$$     and AC = 60 m
Let, AB = h
$${\text{Then,}}\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1$$
$$ ⇒ AB = AC$$
$$ ⇒ h = 60\,m$$
$${\text{And}}\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ AD = \left( {AB \times \sqrt 3 } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 60\sqrt 3 \,m$$
$$\therefore CD = \left( {AD - AC} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\, = 60\left( {\sqrt 3 - 1} \right)\,m$$
$${\text{Hence, required speed }}$$
$${\text{ = }}\left[ {\frac{{60\left( {\sqrt 3 - 1} \right)}}{5}} \right]{\text{m/s}}$$
$${\text{ = }}\left( {12 \times 0.73} \right){\text{m/s}}$$
$$ = \left( {12 \times 0.73 \times \frac{{18}}{5}} \right){\text{km/hr}}$$
$$ = {\text{31}}{\text{.5km/hr }} \approx {\text{ 32km/hr}}$$

Shadow length = $$\frac{1}{{\sqrt 3 }}$$ Height of man
$$\frac{{{\text{Height}}\,{\text{of}}\,{\text{man}}}}{{{\text{Shadow}}\,{\text{length}}}} = \sqrt 3 $$
$$\tan \theta = \frac{{{\text{Height}}\,{\text{of}}\,{\text{man}}}}{{{\text{Shadow}}\,{\text{length}}}} = \sqrt 3 $$
$${\text{But}}\,\tan {60^ \circ } = \sqrt 3 $$
∴ $$\theta $$ = 60° = Angle of elevation of sum

In $$\Delta $$STR, by Pythagoras theorem
$$R{T^2} = S{T^2} + R{S^2}$$
$$\therefore S{T^2} = {15^2} - {9^2} = 144$$
$$\therefore ST = 12\,{\text{m}}$$
In $$\Delta $$TQP, by Pythagoras theorem
$$P{T^2} = T{Q^2} + P{Q^2}$$
$$\therefore T{Q^2} = {15^2} - {12^2} = 81$$
$$\therefore TQ = 9\,{\text{m}}$$
Distance between houses
⇒ SQ = ST + TQ
⇒ SQ = 12 + 9
⇒ SQ = 21 m

Let MN = Pratap's fort & PQ = Suresh's fort
From the diagram we can see that
MR = 24 cm & PR = 15 - 8 = 7 cm
By Pythagoras theorem,
Hypotenuse² = (side1)² + (side2)²
$$\therefore {\text{MP}} = \sqrt {{{24}^2} + {7^2}} $$
$$\therefore {\text{MP}} = 25\,{\text{cm}}$$

$${\text{In}}\,\Delta PBQ,\,\tan {60^ \circ } = \frac{{PQ}}{{BQ}}$$
$$\therefore BQ = \frac{{PQ}}{{\sqrt 3 }}$$
$${\text{In}}\,\Delta PAQ,\,\tan {30^ \circ } = \frac{{PQ}}{{AQ}}$$
$$\therefore \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{50 + BQ}}$$
$$\therefore PQ = \frac{{50 + BQ}}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + \frac{{PQ}}{{\sqrt 3 }}}}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 + PQ}}{{\sqrt 3 \times \sqrt 3 }}$$
$$\therefore 3PQ = 50\sqrt 3 + PQ$$
$$\therefore PQ = 25\sqrt 3 \,{\text{m}} = $$     Height of tree

Angle of Depression = Angle of Elevation
Tower PS = 10 m in height
$${\text{tan}}{45^ \circ } = 1 = \frac{{PS}}{{RS}}$$
$$\therefore PS = RS = 10$$
$$\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PS}}{{SQ}} = \frac{{10}}{{SQ}}$$
$$\therefore SQ = 10\sqrt 3 $$
$$RQ = RS + SQ$$
$$\,\,\,\,\,\,\,\,\,\,\, = 10 + 10\sqrt 3 $$
$$\,\,\,\,\,\,\,\,\,\,\, = 27.32\,{\text{m}}$$