Answer: Option A. -> 30º
Let AB be the tree and AC be its shadow.
Let ∠ACB = θ
$$\frac{{AC}}{{AB}} = \cot \theta $$
$$\frac{{AC}}{{AB}} = \sqrt 3 $$
$$\cot {30^ \circ } = \sqrt 3 $$
∴ θ = 30°

Answer: Option C. -> 27.3 m
Let AB be the tower and C and D be the points of observation.
Then, ∠ACB = 30°, ∠ADB = 45° and CD = 20m
Let AB = h then,
$$\eqalign{
& \frac{{AB}}{{AC}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow AC = AB \times \sqrt 3 = h\sqrt 3 {\kern 1pt} {\text{And,}} \cr
& \Rightarrow \frac{{AB}}{{AD}} = \tan 45^\circ = 1 \cr
& \Rightarrow AD = AB = h \cr
& \, \, \, \, \, CD = 20 \cr
& \Rightarrow \left( {AC - AD} \right) = 20 \cr
& \Rightarrow h\sqrt 3 - h = 20 \cr
& \therefore h = \frac{{20}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr
& = 10\left( {\sqrt 3 + 1} \right){\text{m}} \cr
& = \left( {10 \times 2.73} \right){\text{m}} \cr
& = 27.3 {\text{m}} \cr} $$

Answer: Option C. -> 16 min. 23 sec.
Let AB be the tower and C and D be the two positions of the car.
Then, \[\angle ACB = {45^ \circ },\]    \[\angle ADB = {30^ \circ }\]
Let, AB = h, CD = x and AC = y
$$\eqalign{
& \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr
& \Rightarrow \frac{h}{y} = 1 \cr
& \Rightarrow y = h \cr
& \frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow x + y = \sqrt 3 h \cr
& \therefore x = \left( {x + y - y} \right) \cr
& \,\,\,\,\,\,\, = \sqrt 3 h - h \cr
& \,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right) \cr} $$
Now, $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 12 min.
So, h will be covered in→
$$\eqalign{
& \left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right] \cr
& = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min \cr
& = \left( {\frac{{1200}}{{73}}} \right)\min \cr
& = 16\min ,\,\,23\sec \cr} $$

Answer: Option A. -> 63.5 m
Let AB be the tower and C and D be the objects.
Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
$$\eqalign{
& \frac{{AB}}{{AD}} = \tan {60^ \circ } = \sqrt 3 \cr
& \Rightarrow AD = \frac{{AB}}{{\sqrt 3 }} = \frac{{150}}{{\sqrt 3 }} \cr
& \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr
& \Rightarrow AC = AB = 150{\text{ m}} \cr
& \therefore CD = \left( {AC - AD} \right) \cr
& = \left( {150 - \frac{{150}}{{\sqrt 3 }}} \right){\text{m}} \cr
& = \left[ {\frac{{150\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}} \right]{\text{m}} \cr
& = 50\left( {3 - \sqrt 3 } \right){\text{m}} \cr
& = \left( {50 \times 1.27} \right){\text{m}} \cr
& = 63.5\,{\text{m}} \cr} $$

Answer: Option B. -> \[70\sqrt 3 {\text{ m}}\]
Length of the tower AB = h meter.
$$\eqalign{
& \angle DAC = \angle ACB = {60^ \circ } \cr
& BC = 70{\text{ meter}} \cr
& {\text{In }}\vartriangle {\text{ABC,}} \cr
& {\text{tan }}{60^ \circ } = \frac{{AB}}{{BC}} \cr
& \Rightarrow \sqrt 3 = \frac{h}{{70}} \cr
& \Rightarrow h = 70\sqrt 3 {\text{ meter}} \cr} $$

Answer: Option C. -> 10 meters
Let AB be the tower and CD be the electric pole.
Then, $$\angle ACB = {60^ \circ },$$     $$\angle EDB = {30^ \circ }$$     and AB = 15 m
Let CD = h
Then, BE = (AB - AE) = (AB - CD) = (15 - h)
$$\eqalign{
& \frac{{AB}}{{AC}} = \tan {60^ \circ } = \sqrt 3 \cr
& \Rightarrow AC = \frac{{AB}}{{\sqrt 3 }} = \frac{{15}}{{\sqrt 3 }} \cr
& {\text{And, }}\frac{{BE}}{{DE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow DE = \left( {BE \times \sqrt 3 } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {15 - h} \right) \cr
& {\text{So, }}AC = DE \cr
& \Rightarrow \frac{{15}}{{\sqrt 3 }} = \sqrt 3 \left( {15 - h} \right) \cr
& \Rightarrow 3h = \left( {45 - 15} \right) \cr
& \Rightarrow h = 10{\text{ m}} \cr} $$

Answer: Option B. -> 45°
$${\text{Given tan }}{x^ \circ } = \frac{2}{5}$$     and AF = 200 meter
$$\eqalign{
& \therefore \frac{2}{5} = \frac{{TF}}{{AF}} \cr
& \Rightarrow TF = \frac{{2 \times 200}}{5} \cr
& \Rightarrow TF = 80{\text{ m}} \cr
& {\text{We have, BF = 80 m}} \cr
& \therefore \tan {\text{ }}{y^ \circ } = \frac{{TF}}{{BF}} \cr
& \Rightarrow \tan {\text{ }}{y^ \circ } = \frac{{80}}{{80}} \cr
& \Rightarrow \tan {\text{ }}{y^ \circ } = 1 = \tan {45^ \circ } \cr
& \therefore {y^ \circ } = {45^ \circ } \cr} $$

Answer: Option A. -> $$\left( {\sqrt 3 + 1} \right)\,h\,{\text{metres}}$$
Let AB be lighthouse and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB = x and QB = y
$$\eqalign{
& {\text{Now in right }}\Delta APB \cr
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr
& \Rightarrow \tan {45^ \circ } = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \cr
& \Rightarrow x = h\,............(i) \cr
& {\text{Similarly in right }}\Delta AQB, \cr
& \tan {30^ \circ } = \frac{{AP}}{{QB}} = \frac{h}{y} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{y} \cr
& \Rightarrow y = \sqrt 3 \,h\,..............(ii) \cr
& {\text{Adding (i) and (ii)}} \cr
& \therefore PQ = x + y \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = h + \sqrt 3 \,h \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt 3 + 1} \right)\,h \cr} $$

Answer: Option C. -> 300 m
$$\eqalign{
& {\text{Let, }}OP = {\text{ }}a \cr
& {\text{tan }}{60^ \circ } = \frac{H}{a} \cr
& \Rightarrow H = \sqrt 3 a \cr
& \Rightarrow \frac{H}{{\sqrt 3 }} = a.....(i) \cr} $$
$$tan{45^ \circ } = 1$$   $$ = \frac{H}{{a + 100\left( {3 - \sqrt 3 } \right)}}$$
$$ \Rightarrow a + 100\left( {3 - \sqrt 3 } \right) = H$$
From (i) $$\frac{H}{{\sqrt 3 }} + $$   $$100\left( {3 - \sqrt 3 } \right)$$   = H
$$\eqalign{
& \Rightarrow H + 300\sqrt 3 - 300 = \sqrt 3 H \cr
& \Rightarrow 300\sqrt 3 - 300 = \sqrt 3 H - H \cr
& \Rightarrow \left( {\sqrt 3 - 1} \right)H = 300\left( {\sqrt 3 - 1} \right) \cr
& \Rightarrow H = 300{\text{ m}} \cr} $$

Answer: Option C. -> 2.5 m
Let AB is girls and CD is lamp-post AB = 1.5
which casts her shadow EB
$$\eqalign{
& \therefore EB = 4.5\,m,\,BD = 3\,m \cr
& {\text{Now in }}\Delta AEB \cr
& \tan \theta = \frac{{AB}}{{BE}} = \frac{{1.5}}{{4.5}} = \frac{1}{3} \cr
& {\text{and in }}\Delta CED \cr
& \tan \theta = \frac{{CD}}{{ED}} \Rightarrow \frac{1}{3} = \frac{h}{{4.5 + 3}} \cr
& \Rightarrow \frac{1}{3} = \frac{h}{{7.5}} \cr
& \Rightarrow h = \frac{{7.5}}{3} - 2.5\,m \cr
& \therefore {\text{Height of lamp - post}} = {\text{2}}{\text{.5}}\,m \cr} $$