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.)

Let AB be rod and BC be it's shadow
So that AB : BC = 1 : $$\sqrt 3 $$
Let $$\theta $$ be the angle of elevation

$$\therefore \tan \theta = \frac{{AB}}{{BC}} = \frac{1}{{\sqrt 3 }} = \tan {30^ \circ }$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \tan {{30}^ \circ } = \frac{1}{{\sqrt 3 }}} \right)$$
$$\therefore \theta = {30^ \circ }$$
∴ Hence angle of elevation $$ = {30^ \circ }$$

In the figure, AB is chimney and CB and DB are it's shadow

$$\tan {60^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{{BC}}$$
$$ ⇒ \sqrt 3 = \frac{h}{{BC}}$$
$$ ⇒ BC = \frac{h}{{\sqrt 3 }}\,.......\,\left( {\text{i}} \right)$$
$${\text{and}}$$
$$\tan {30^ \circ } = \frac{h}{{DB}} = \frac{h}{{DB + BC}}$$
$$\frac{1}{{\sqrt 3 }} = \frac{h}{{x + BC}} \cr &×+ BC = h\sqrt 3 $$
$$ ⇒ BC = h\sqrt 3 - x\,.......\,\left( {{\text{ii}}} \right)$$
$${\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right)$$
$$\frac{h}{{\sqrt 3 }} = h\sqrt 3 -× $$
$$ ⇒ \frac{h}{{\sqrt 3 }} - h\sqrt 3 = -× \cr &×= h\sqrt 3 - \frac{h}{{\sqrt 3 }} \cr &×= h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) \cr &×= h\frac{{3 - 1}}{{\sqrt 3 }} \cr &×= \frac{{2h}}{{\sqrt 3 }}$$
$$\therefore h = \frac{{\sqrt 3 }}{2}x$$

Suppose AB is the ladder of length×m
∴ OA = 2m, ∠OAB = 60°

$${\text{In right }}\Delta AOB,\,\sec {60^ \circ } = \frac{x}{2}$$
$$ ⇒ 2 = \frac{x}{2}$$
$$ ⇒ x = 4\,m$$

Let AB and CD be two poles
AB = 20 m, CD = 14 m
A and C are joined by a wire
CE || DB and angle of elevation of A is 30°
Let CE = DB =×and AC = L

Now AE = AB - EB = AB - CD = 20 - 14 = 6 m
$${\text{Now in right }}\Delta ACE,$$
$$\sin \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} = \frac{{AE}}{{AC}}$$
$$ ⇒ \sin {30^ \circ } = \frac{6}{{AC}}$$
$$ ⇒ \frac{1}{2} = \frac{6}{{AC}}$$
$$ ⇒ AC = 2 \times 6 = 12$$
$$\therefore {\text{Length of AC}} = 12\,m$$

Let CD is the tower and A is a point such that the angle of elevation of C is 30°
B is and their point h m high of A and angle of depression of D is 60°

$${\text{The}}\,AB = h\,m$$
$${\text{Let}}\,CD = H\,m\,\,\,{\text{and }}\,AD\, =× $$
$$ {\text{Now}}\,{\text{in}}\,{\text{right}}\,\Delta ABD,$$
$$\tan \theta = \frac{{AB}}{{AD}}$$
$$ ⇒ \tan {60^ \circ } = \frac{h}{x}$$
$$ ⇒ \sqrt 3 = \frac{h}{x}$$
$$ ⇒ x = \frac{h}{{\sqrt 3 }}\,.......({\text{i}})$$
$${\text{Similarly in right }}\Delta ACB,$$
$$\tan {30^ \circ } = \frac{{CD}}{{AD}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{H}{x}$$
$$ ⇒ x = \sqrt 3 \,H\,.......\left( {{\text{ii}}} \right)$$
$${\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right)$$
$$\sqrt 3 \,H = \frac{h}{{\sqrt 3 }}$$
$$H = \frac{h}{{\sqrt 3 \times \sqrt 3 }} = \frac{h}{3}$$
$$\therefore {\text{Height of tower}} = \frac{h}{3}$$