Answer: Option B. -> $$\frac{a}{{2\sqrt 2 }}\,{\text{metres}}$$
Let height of pole CD = h
and AB = 2h, BD = a
M is mid-point of BD
$$\therefore DM = MB = \frac{a}{2}$$
$${\text{Let }}\angle CMD = \theta ,$$    $${\text{then }}\angle AMB = $$     $${90^ \circ } - \theta $$
$$\eqalign{
& {\text{Now}} \cr
& \tan \theta = \frac{{CD}}{{DM}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a}\,......({\text{i}}) \cr
& {\text{and}} \cr
& tan\left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{MB}} = \frac{{2h}}{{\frac{a}{2}}} = \frac{{4h}}{a} \cr
& \Rightarrow \cot \theta = \frac{{4h}}{a}\,..........({\text{ii}}) \cr
& {\text{Multiplying (i) and (ii)}} \cr
& {\text{tan}}\theta \times {\text{cot}}\theta = \frac{{2h}}{a} \times \frac{{4h}}{a} \cr
& 1 = \frac{{8{h^2}}}{{{a^2}}} = {h^2} = \frac{{{a^2}}}{8}\,m \cr
& h = \sqrt {\frac{{{a^2}}}{8}} = \frac{a}{{\sqrt 8 }} = \frac{a}{{2\sqrt 2 }}\,m \cr} $$

Answer: Option B. -> 50 m
Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C
Let angle be Q and CD = 25 m
$$\eqalign{
& {\text{Let}}\,AB = h \cr
& CE \,\, || \,\, DB \cr
& \therefore EC = DB = x{\text{ }}\left( {{\text{suppose}}} \right) \cr
& EB = CD = 25 \cr
& \therefore AE = h - 25 \cr
& {\text{Now in right }}\Delta CDB, \cr
& \tan \theta = \frac{{CD}}{{DB}} = \frac{{25}}{x}\,......\left( {\text{i}} \right) \cr
& {\text{and in right }}\Delta CAE \cr
& \tan \theta = \frac{{AE}}{{CE}} = \frac{{h - 25}}{x}\,......\left( {{\text{ii}}} \right) \cr
& {\text{From}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr
& \frac{{25}}{x} = \frac{{h - 25}}{x} \cr
& \Rightarrow 25 = h - 25 \cr
& \Rightarrow h = 25 + 25 = 50 \cr
& \therefore {\text{Height of tower}} = 50\,m \cr} $$

Answer: Option B. -> $$50$$
Let AB be tower and C is a point on the ground 50 m away
From foot of tower B
Angle of elevation is 45°
Let h be height of tower = x m
$$\eqalign{
& \therefore \tan \theta = \frac{{AB}}{{BC}} \cr
& \Rightarrow \tan {45^ \circ } = \frac{{AB}}{50} \cr
& \Rightarrow 1 = \frac{{AB}}{{50}} \Rightarrow AB = 50\,m \cr} $$

Answer: Option C. -> $$25\sqrt 3 \,{\text{m}}$$
$$\eqalign{
& {\text{In}}\,\Delta PBQ,\,\tan {60^ \circ } = \frac{{PQ}}{{BQ}} \cr
& \therefore BQ = \frac{{PQ}}{{\sqrt 3 }} \cr
& {\text{In}}\,\Delta PAQ,\,\tan {30^ \circ } = \frac{{PQ}}{{AQ}} \cr
& \therefore \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{50 + BQ}} \cr
& \therefore PQ = \frac{{50 + BQ}}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + \frac{{PQ}}{{\sqrt 3 }}}}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 + PQ}}{{\sqrt 3 \times \sqrt 3 }} \cr
& \therefore 3PQ = 50\sqrt 3 + PQ \cr} $$
$$\therefore PQ = 25\sqrt 3 \,{\text{m}} = $$     Height of tree

Answer: Option D. -> 60°
Shadow length = $$\frac{1}{{\sqrt 3 }}$$ Height of man
$$\eqalign{
& \frac{{{\text{Height}}\,{\text{of}}\,{\text{man}}}}{{{\text{Shadow}}\,{\text{length}}}} = \sqrt 3 \cr
& \tan \theta = \frac{{{\text{Height}}\,{\text{of}}\,{\text{man}}}}{{{\text{Shadow}}\,{\text{length}}}} = \sqrt 3 \cr
& {\text{But}}\,\tan {60^ \circ } = \sqrt 3 \cr} $$
∴ $$\theta $$ = 60° = Angle of elevation of sum

Answer: Option A. -> 27.32 m
Angle of Depression = Angle of Elevation
Tower PS = 10 m in height
$$\eqalign{
& {\text{tan}}{45^ \circ } = 1 = \frac{{PS}}{{RS}} \cr
& \therefore PS = RS = 10 \cr
& \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PS}}{{SQ}} = \frac{{10}}{{SQ}} \cr
& \therefore SQ = 10\sqrt 3 \cr
& RQ = RS + SQ \cr
& \,\,\,\,\,\,\,\,\,\,\, = 10 + 10\sqrt 3 \cr
& \,\,\,\,\,\,\,\,\,\,\, = 27.32\,{\text{m}} \cr} $$

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

Answer: Option C. -> 25 cm
Let MN = Ramesh'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
$$\eqalign{
& \therefore {\text{MP}} = \sqrt {{{24}^2} + {7^2}} \cr
& \therefore {\text{MP}} = 25\,{\text{cm}} \cr} $$

Answer: Option A. -> $$30\sqrt 3 \,m$$
Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60°
$$\eqalign{
& {\text{Let height of tower }}AB = h \cr
& {\text{Then in right }}\Delta APB, \cr
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr
& \Rightarrow \tan {60^ \circ } = \frac{h}{{30}} \cr
& \Rightarrow \sqrt 3 = \frac{h}{{30}} \cr
& \Rightarrow h = 30\sqrt 3 \cr} $$
∴ Height of the tower $$ = 30\sqrt 3 \,m$$

Answer: Option C. -> $$100\left( {\sqrt 3 - 1} \right)\,m$$
Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y
$$\eqalign{
& {\text{Now in right }}\Delta ADB, \cr
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{DB}} \cr
& \tan {45^ \circ } = \frac{{100}}{y} \cr
& \Rightarrow 1 = \frac{{100}}{y} \cr
& \Rightarrow y = 100 \cr
& {\text{Similarly in right }}\Delta ACB, \cr
& \tan {30^ \circ } = \frac{{AB}}{{CB}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{y + x}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{100 + x}} \cr
& \Rightarrow 100 + x = 100\sqrt 3 \cr
& \Rightarrow x = 100\sqrt 3 - 100 \cr
& \Rightarrow x = 100\left( {\sqrt 3 - 1} \right)\,m \cr} $$