Answer: Option A. -> 30°
Let AB be rod and BC be its shadow
So that AB : BC = 1 : $$\sqrt 3 $$
Let $$\theta $$ be the angle of elevation
$$\eqalign{
& \therefore \tan \theta = \frac{{AB}}{{BC}} = \frac{1}{{\sqrt 3 }} = \tan {30^ \circ } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \tan {{30}^ \circ } = \frac{1}{{\sqrt 3 }}} \right) \cr
& \therefore \theta = {30^ \circ } \cr} $$
∴ Hence angle of elevation $$ = {30^ \circ }$$

Answer: Option C. -> $$\frac{h}{3}\,m$$
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°
$$\eqalign{
& {\text{The}}\,AB = h\,m \cr
& {\text{Let}}\,CD = H\,m\,\,\,{\text{and }}\,AD\, = x \cr
& {\text{Now}}\,{\text{in}}\,{\text{right}}\,\Delta ABD, \cr
& \tan \theta = \frac{{AB}}{{AD}} \cr
& \Rightarrow \tan {60^ \circ } = \frac{h}{x} \cr
& \Rightarrow \sqrt 3 = \frac{h}{x} \cr
& \Rightarrow x = \frac{h}{{\sqrt 3 }}\,.......({\text{i}}) \cr
& {\text{Similarly in right }}\Delta ACB, \cr
& \tan {30^ \circ } = \frac{{CD}}{{AD}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{H}{x} \cr
& \Rightarrow x = \sqrt 3 \,H\,.......\left( {{\text{ii}}} \right) \cr
& {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr
& \sqrt 3 \,H = \frac{h}{{\sqrt 3 }} \cr
& H = \frac{h}{{\sqrt 3 \times \sqrt 3 }} = \frac{h}{3} \cr
& \therefore {\text{Height of tower}} = \frac{h}{3} \cr} $$

Answer: Option A. -> 12 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 = x and AC = L
Now AE = AB - EB = AB - CD = 20 - 14 = 6 m
$$\eqalign{
& {\text{Now in right }}\Delta ACE, \cr
& \sin \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} = \frac{{AE}}{{AC}} \cr
& \Rightarrow \sin {30^ \circ } = \frac{6}{{AC}} \cr
& \Rightarrow \frac{1}{2} = \frac{6}{{AC}} \cr
& \Rightarrow AC = 2 \times 6 = 12 \cr
& \therefore {\text{Length of AC}} = 12\,m \cr} $$

Answer: Option D. -> $$4$$
Suppose AB is the ladder of length x m
∴ OA = 2m, ∠OAB = 60°
$$\eqalign{
& {\text{In right }}\Delta AOB,\,\sec {60^ \circ } = \frac{x}{2} \cr
& \Rightarrow 2 = \frac{x}{2} \cr
& \Rightarrow x = 4\,m \cr} $$

Answer: Option C. -> $$\frac{{\sqrt 3 }}{2}\,x$$
In the figure, AB is chimney and CB and DB are its shadow
$$\eqalign{
& \tan {60^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{{BC}} \cr
& \Rightarrow \sqrt 3 = \frac{h}{{BC}} \cr
& \Rightarrow BC = \frac{h}{{\sqrt 3 }}\,.......\,\left( {\text{i}} \right) \cr
& {\text{and}} \cr
& \tan {30^ \circ } = \frac{h}{{DB}} = \frac{h}{{DB + BC}} \cr
& \frac{1}{{\sqrt 3 }} = \frac{h}{{x + BC}} \cr
& x + BC = h\sqrt 3 \cr
& \Rightarrow BC = h\sqrt 3 - x\,.......\,\left( {{\text{ii}}} \right) \cr
& {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr
& \frac{h}{{\sqrt 3 }} = h\sqrt 3 - x \cr
& \Rightarrow \frac{h}{{\sqrt 3 }} - h\sqrt 3 = - x \cr
& x = h\sqrt 3 - \frac{h}{{\sqrt 3 }} \cr
& x = h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) \cr
& x = h\frac{{3 - 1}}{{\sqrt 3 }} \cr
& x = \frac{{2h}}{{\sqrt 3 }} \cr
& \therefore h = \frac{{\sqrt 3 }}{2}x \cr} $$

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

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

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

Answer: Option B. -> 8 cm
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
$$\eqalign{
& {\text{in}}\,\Delta MCQ, \cr
& \sin {30^ \circ } = \frac{1}{2} = \frac{{MQ}}{{MN}} = \frac{{\text{h}}}{{24 - {\text{h}}}} \cr
& \therefore 24 - {\text{h}} = 2{\text{h}} \cr} $$
∴ h = 8 cm = Tree breaks at this height

Answer: Option D. -> 15 m
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