Answer: Option D. -> $$mn$$
$$\eqalign{
& {\text{tan }}\theta = \frac{H}{{{m^2}}} \cr
& \Rightarrow \tan \left( {90^ \circ - \theta } \right) = \frac{H}{{{n^2}}} \cr
& \Rightarrow \cot \theta = \frac{H}{{{n^2}}} \cr
& \Rightarrow \tan \theta .\cot \theta = \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} \cr
& \Rightarrow \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} = 1 \cr
& \Rightarrow {H^2} = {m^2}{n^2} \cr
& \Rightarrow H = mn \cr} $$

Answer: Option A. -> Equal
$$\eqalign{
& {\text{Here, AD is parallel to BC and AC is a transversal}}{\text{.}} \cr
& \Rightarrow {\theta _1} = {\theta _2} \cr} $$

Answer: Option A. -> 32 kmph
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
$$\eqalign{
& {\text{Then,}}\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr
& \Rightarrow AB = AC \cr
& \Rightarrow h = 60\,m \cr
& {\text{And}}\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow AD = \left( {AB \times \sqrt 3 } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 60\sqrt 3 \,m \cr
& \therefore CD = \left( {AD - AC} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\, = 60\left( {\sqrt 3 - 1} \right)\,m \cr
& {\text{Hence, required speed }} \cr
& {\text{ = }}\left[ {\frac{{60\left( {\sqrt 3 - 1} \right)}}{5}} \right]{\text{m/s}} \cr
& {\text{ = }}\left( {12 \times 0.73} \right){\text{m/s}} \cr
& = \left( {12 \times 0.73 \times \frac{{18}}{5}} \right){\text{km/hr}} \cr
& = {\text{31}}{\text{.5km/hr }} \approx {\text{ 32km/hr}} \cr} $$

Answer: Option B. -> 60°
Let AB be a vertical pole and let its shadow be BC
Let BC = x m, then length of pole = $$\sqrt 3 $$ x,
$$\theta $$ be the angle of elevation
$$\eqalign{
& \therefore {\text{tan}}\theta = \frac{{AB}}{{BC}} = \frac{{\sqrt 3 \,x}}{x} = \sqrt 3 \cr
& = \tan {60^ \circ } \cr
& \therefore \theta = {60^ \circ } \cr} $$

Answer: Option A. -> $$100\sqrt 3 \,m$$
Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60°
Let height of the tower = h
$$\eqalign{
& {\text{Then in right }}\Delta ABP \cr
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr
& \Rightarrow \tan {60^ \circ } = \frac{h}{{100}} \Rightarrow \sqrt 3 = \frac{h}{{100}} \cr
& \Rightarrow h = 100\sqrt 3 \cr
& \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = 100\sqrt 3 \cr} $$

Answer: Option C. -> $$50\left( {\sqrt 3 - 1} \right)m$$
Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively
∠ACB = ∠XAC = 30° , ∠ADB = ∠YAD = 45° and
CD = 100m
Let AB =h and CB = x then BC = (100 - x)m
$$\eqalign{
& {\text{Now in }}\,\Delta ACB, \cr
& \tan \theta = \frac{{AB}}{{CB}} \cr
& \tan {30^ \circ } = \frac{h}{x} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr
& \Rightarrow x = \sqrt 3 h ......{\text{(i)}} \cr
& {\text{Similarly in }}\,\Delta ADB \cr
& \tan {45^ \circ } = \frac{{AB}}{{BD}} \cr
& \Rightarrow 1 = \frac{h}{{100 - x}} \cr
& \Rightarrow x = 100 - h \,......{\text{(ii)}} \cr
& {\text{From (i) and (ii)}} \cr
& \sqrt 3 h = 100 - h \cr
& \Rightarrow (\sqrt 3 - 1)h = 100 \cr
& h = \frac{{100}}{{\sqrt 3 + 1}} \cr
& \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} \cr
& \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} \cr
& \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{2} \cr
& \,\,\,\,\,\, = 50\left( {\sqrt 3 - 1} \right) \cr} $$
∴ height of light house $$ = 50\left( {\sqrt 3 - 1} \right)m$$

Answer: Option D. -> $$\frac{a}{{2\sqrt 2 }}$$
Let AB and CD are two persons standing ‘a’ meters apart
P is the mid-point of BD and from M, the angles of elevation of A and C are complementary
$$\eqalign{
& {\text{In}}\,\,\Delta {\text{APB,}} \cr
& \tan \theta = \frac{{AB}}{{BP}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a} \cr
& {\text{In}}\,\,\Delta {\text{CDP,}} \cr
& \cot (90 - \theta ) = \frac{{PD}}{{CD}} = \frac{{\frac{a}{2}}}{{2h}} = \frac{a}{{4h}} \cr
& {\text{We}}\,\,{\text{Know}}\,\,{\text{that,}} \cr
& \tan \theta = \cot (90 - \theta ). \cr
& \therefore \frac{{2h}}{a} = \frac{a}{{4h}} \cr
& \Rightarrow 8{h^2} = {a^2} \cr
& \Rightarrow h = \frac{a}{{2\sqrt 2 }} \cr} $$

Answer: Option C. -> 75 $$\sqrt 3 $$
AB is a tower and AB = 75 m
From A, the angle of depression of a car C
on the ground is 30°
$$\eqalign{
& {\text{Let distance }}BC = x \cr
& {\text{Now in right }}\Delta ACB, \cr
& \tan \theta = \frac{{AB}}{{BC}} \cr
& \Rightarrow \tan {30^ \circ } = \frac{{75}}{x} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{75}}{x} \cr
& \Rightarrow x = 75\sqrt 3 \,m \cr
& \therefore BC = 75\sqrt 3 \,m \cr} $$

Answer: Option B. -> $$\sqrt {ab} $$
Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively
$$\eqalign{
& {\text{Let }}AB = h \cr
& {\text{Now in right }}\Delta APB, \cr
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr
& \Rightarrow \tan {30^ \circ } = \frac{h}{a} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{a}\,...........(i) \cr
& {\text{Similarly in right }}\Delta AQB, \cr
& \tan {60^ \circ } = \frac{{AB}}{{QB}} \cr
& \Rightarrow \sqrt 3 = \frac{h}{b}\,...........(ii) \cr
& {\text{Multiplying (i) and (ii)}} \cr
& \frac{1}{{\sqrt 3 }} \times \sqrt 3 = \frac{h}{a} \times \frac{h}{b} \cr
& \Rightarrow 1 = \frac{{{h^2}}}{{ab}} \cr
& \Rightarrow {h^2} = ab \cr
& \Rightarrow h = \sqrt {ab} \cr
& \therefore {\text{Height of the tower}} = \sqrt {ab} \cr} $$

Answer: Option B. -> $$\frac{{15\sqrt 3 }}{2}\,m$$
Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground
∴ In ∆ABC, ∠B = 90°
Let height of wall AB = h
Then
$$\sin \theta = \frac{{AB}}{{AC}} \Rightarrow \sin {60^ \circ } = \frac{h}{{15}}$$
$$\eqalign{
& \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{h}{{15}} \cr
& \Rightarrow h = \frac{{15\sqrt 3 }}{2}\,m \cr} $$
∴ Height of the wall $$ = \frac{{15\sqrt 3 }}{2}\,m$$