Height and Distance

$$\tan {30^ \circ } = \frac{{RQ}}{{PQ}}$$
$$\frac{1}{{\sqrt 3 }} = \frac{{200}}{{PQ}}$$
$$PQ = 200\sqrt 3 $$
$$\,\,\,\,\,\,\,\,\,\, = 200 \times 1.73$$
$$\,\,\,\,\,\,\,\,\,\, = 346\,{\text{m}}$$

Let DC be the wall, AB be the tree.
Given that ∠DBC = 30°, ∠DAE = 60°, DC = 11 m
$$\tan {30^ \circ } = \frac{{DC}}{{BC}}$$
$$\frac{1}{{\sqrt 3 }} = \frac{{11}}{{BC}}$$
$$BC = 11\sqrt 3 \,m$$
$$AE = BC = 11\sqrt 3 \,m\,.....\left( 1 \right)$$
$$\tan {60^ \circ } = \frac{{ED}}{{AE}}$$
$$\sqrt 3 = \frac{{ED}}{{11\sqrt 3 }}$$   [∵ Substituted value of AE from (1)]
$$ED = 11\sqrt 3 \times \sqrt 3 $$
$$\,\,\,\,\,\,\,\,\,\, = 11 \times 3$$
$$\,\,\,\,\,\,\,\,\,\, = 33$$
$${\text{Height}}\,{\text{of}}\,{\text{the}}\,{\text{tree}}$$
$$ = AB = EC = \left( {ED + DC} \right)$$
$$ = 33 + 11$$
$$ = 44\,{\text{m}}$$

Consider a right-angled triangle PQR as shown below.

Let QR = 3 and PR = 5 such that
$$\sin \theta = \frac{3}{5}\,\,\,\,\left[ {{\text{i}}{\text{.e}}{\text{.}},\theta = {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right]$$
$$PQ = \sqrt {P{R^2} - Q{R^2}} $$     (∵ Pythagorean theorem)
$$ = \sqrt {{5^2} - {3^2}} $$
$$ = 4$$
$${\text{i}}{\text{.e}}{\text{.}},\,{\text{when}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right),$$     PQ : QR = 4 : 3 ......(eq : 1)

Now Let's solve the question. Let P be the point of observation and QR be the tower as shown in the below diagram.

$${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$     and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq : 1)
$${\text{i}}{\text{.e}}{\text{.}},\,20:QR = 4:3$$
$$ ⇒ 20 \times 3 = QR \times 4$$
$$ ⇒ QR = 15\,{\text{m}}$$
i.e. height of the tower = 15 m

Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC
Given that BA = 10 m, ∠ BAC = 60°
$$\sin {60^ \circ } = \frac{{BC}}{{BA}}$$
$$\frac{{\sqrt 3 }}{2} = \frac{{BC}}{{10}}$$
$$BC = 10 \times \frac{{\sqrt 3 }}{2}$$
$$\,\,\,\,\,\,\,\,\,\, = 5 \times 1.73$$
$$\,\,\,\,\,\,\,\,\,\, = 8.65\,{\text{m}}$$

$$\tan {45^ \circ } = \frac{{SR}}{{QR}}$$
$$\tan {30^ \circ } = \frac{{SR}}{{PR}} = \frac{{SR}}{{\left( {PQ + QR} \right)}}$$
Two equations and 3 variables. Hence we can not find the required value with the given data.

(Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value.)

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

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

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

$${\text{Let, }}AC = H{\text{ }}m$$
$$OA = {\text{ }}a{\text{ }}m$$
$$\tan {\text{ }}{30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{H}{a}$$
$$\tan {60^ \circ } = \sqrt 3 $$    $$ = \frac{{H + 200 + 200}}{a}$$
$$ ⇒ \frac{{H + 400}}{{\sqrt 3 H}} = \sqrt 3 $$
$$ ⇒ H + 400 = 3H$$
$$ ⇒ 2H = 400\,m$$
$$ ⇒ H = 200\,m$$
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
$${\text{Then,}}\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1$$
$$ ⇒ AB = AC$$
$$ ⇒ h = 60\,m$$
$${\text{And}}\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ ⇒ AD = \left( {AB \times \sqrt 3 } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 60\sqrt 3 \,m$$
$$\therefore CD = \left( {AD - AC} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\, = 60\left( {\sqrt 3 - 1} \right)\,m$$
$${\text{Hence, required speed }}$$
$${\text{ = }}\left[ {\frac{{60\left( {\sqrt 3 - 1} \right)}}{5}} \right]{\text{m/s}}$$
$${\text{ = }}\left( {12 \times 0.73} \right){\text{m/s}}$$
$$ = \left( {12 \times 0.73 \times \frac{{18}}{5}} \right){\text{km/hr}}$$
$$ = {\text{31}}{\text{.5km/hr }} \approx {\text{ 32km/hr}}$$