Answer: Option C. -> 12
Let AB and CD are two poles AB = 10 m and CD = 16 m
AC is wire which makes an angle of 30° with the horizontal
Let BD = x, then AE = x
CE = CD - ED = CD - AB = 16 - 10 = 6m
$$\eqalign{
& {\text{Now}}\,{\text{in}}\,\Delta ACE \cr
& \sin {30^ \circ } = \frac{{CE}}{{AC}} = \frac{6}{l} \cr
& \Rightarrow \frac{1}{2} = \frac{6}{l} \cr
& \Rightarrow l = 2 \times 6 = 12\,m \cr} $$

Answer: Option B. -> 150 $$\sqrt 3 $$
Let AB be the tower of height 150 m
C is car and angle of depression is 30°
Therefore, ∠ACB = 30° (alternate angle)
In right - angled triangle ABC,
$$\eqalign{
& \frac{{BC}}{{AB}} = \cot {30^ \circ } \cr
& \Rightarrow \frac{{BC}}{{150}} = \sqrt 3 \cr
& \Rightarrow BC = 150\sqrt 3 \,m \cr} $$
That is, distance of the car from the tower is $$150\sqrt 3 \,m$$

Answer: Option B. -> $$\frac{d}{{\cot \alpha - \cot \beta }}$$
Let AB be the tower and C is a point such that the angle of elevation of A is α.
After walking towards the foot B of the tower, at D the angle of elevation is β.
Let h be the height of the tower and DB = x Now in ΔACB,
$$\eqalign{
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{CB}} \cr
& \tan \alpha = \frac{h}{{d + x}} \cr
& \Rightarrow d + x = \frac{h}{{\tan \alpha }} \cr
& \Rightarrow d + x = h\cot \alpha \cr
& \Rightarrow x = h\cot \alpha - d\,.......\left( {\text{i}} \right) \cr
& {\text{Similarly in right }}\Delta ADB, \cr
& \tan \beta = \frac{h}{x} \cr
& \Rightarrow x = \frac{h}{{\tan \beta }} \cr
& \Rightarrow x = h\cot \beta \,.........({\text{ii}}) \cr
& {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr
& h\cot \alpha - d = h\cot \beta \cr
& \Rightarrow h\cot \alpha - h\cot \beta = d \cr
& \Rightarrow h\left( {\cot \alpha - \cot \beta } \right) = d \cr
& \Rightarrow h = \frac{d}{{\cot \alpha - \cot \beta }} \cr} $$
∴ Height of the tower $$ = \frac{d}{{\cot \alpha - \cot \beta }}$$

Answer: Option B. -> 30°
Let AB be tower and BC be its shadow
∴ Let AB = x
$$\eqalign{
& {\text{Then}}\,BC = \sqrt 3 \times x = \sqrt 3 \,x \cr
& \therefore \tan \theta = \frac{{AB}}{{BC}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{{\sqrt 3 \,x}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \tan {30^ \circ } \cr
& \therefore \theta = {30^ \circ } \cr} $$
∴ Angle of elevation of the sun$${\text{ = }}{30^ \circ }$$

Answer: Option A. -> $$\left( {\sqrt 3 + 1} \right)\,x$$
AB is a tower
BD and BC are its shadows and CD = 2x
$$\eqalign{
& \tan {45^ \circ } = \frac{{AB}}{{DB}} \cr
& \Rightarrow 1 = \frac{h}{y} \Rightarrow y = h \cr
& {\text{and}}\tan {30^ \circ } = \frac{{AB}}{{CB}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{2x + y}} \cr
& \Rightarrow 2x + y = \sqrt 3 \,h \cr
& \Rightarrow \sqrt 3 \,h - h = 2x \cr
& \Rightarrow h\left( {\sqrt 3 - 1} \right) = 2x \cr
& \Rightarrow h = \frac{{2x}}{{\sqrt 3 - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {\sqrt 3 + 1} \right) \cr} $$

Answer: Option B. -> $$\sqrt {ab} $$
Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are $$\theta $$ and (90° – $$\theta $$)
Let height of tower = h
$$\eqalign{
& {\text{Then}}\,{\text{in}}\,{\text{right}}\,\Delta APB, \cr
& \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr
& = \frac{h}{a}\,............\left( {\text{i}} \right) \cr
& {\text{Similarly in right }}\Delta AQB, \cr
& \tan \left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{QB}} = \frac{h}{b} \cr
& \Rightarrow \cot \theta = \frac{h}{b}\,...........\left( {{\text{ii}}} \right) \cr
& {\text{Multiplying }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr
& \tan \theta \cot \theta = \frac{h}{a} \times \frac{h}{b} \cr
& \Rightarrow 1 = \frac{{{h^2}}}{{ab}} \cr
& \Rightarrow {h^2} = ab \cr
& \Rightarrow h = \sqrt {ab} \cr
& \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = \sqrt {ab} \cr} $$

Answer: Option D. -> 254 m
Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m, ∠DAC = 45°, ∠DBC = 60°
$$\eqalign{
& \tan {60^ \circ } = \frac{{DC}}{{BC}} \cr
& \sqrt 3 = \frac{{600}}{{BC}} \cr
& BC = \frac{{600}}{{\sqrt 3 }}\,..........\,\left( {\text{i}} \right) \cr
& \tan {45^ \circ } = \frac{{DC}}{{AC}} \cr
& 1 = \frac{{600}}{{AC}} \cr
& AC = 600\,............\,\left( {{\text{ii}}} \right) \cr} $$
Distance between the objects
$$ = AC = \left( {AC - BC} \right)$$
$$ = 600 - \frac{{600}}{{\sqrt 3 }}$$   [∵ from (1) and (ii)]
$$\eqalign{
& = 600\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr
& = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \cr
& = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr
& = \frac{{600\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{3} \cr
& = 200\sqrt 3 \left( {\sqrt 3 - 1} \right) \cr
& = 200\left( {3 - \sqrt 3 } \right) \cr
& = 200\left( {3 - 1.73} \right) \cr
& = 254\,m \cr} $$

Answer: Option D. -> 94.6 m
Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°, ∠BCD = 60°
$$\eqalign{
& \tan {45^ \circ } = \frac{{BD}}{{BA}} \cr
& \Rightarrow 1 = \frac{{60}}{{BA}} \cr
& \Rightarrow BA = 60\,m\,........\left( {\text{i}} \right) \cr
& \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr
& \Rightarrow \sqrt 3 = \frac{{60}}{{BC}} \cr
& \Rightarrow BC = \frac{{60}}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60\sqrt 3 }}{3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\sqrt 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 1.73 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.6\,m\,..........\left( {{\text{ii}}} \right) \cr} $$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m

Answer: Option D. -> 800 sq.ft.
$$\eqalign{
& \tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{QR}} \cr
& \therefore PQ = \sqrt 3 \,QR \cr
& \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{SQ}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{80 + QR}} \cr
& \therefore 80 + QR = \sqrt 3 \,PQ \cr
& \therefore 80 + QR = 3QR \cr
& \therefore QR = 40\,{\text{ft}}{\text{.}} \cr} $$
If we read carefully, we see that Raj (Point R) and the scarecrow (Point Q) are in diagonally opposite corners.
So QR is a diagonal of the square farm.
Diagonal of square = side x$$\sqrt 2 $$
$$\eqalign{
& \therefore 40 = {\text{side}}\,{\text{x}}\sqrt 2 \cr
& \therefore {\text{side}} = \frac{{40}}{{\sqrt 2 }} \cr
& \therefore {\text{Area}} = {\left( {{\text{side}}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{{40}}{{\sqrt 2 }}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{1600}}{2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 800\,{\text{sq}}{\text{.}}\,{\text{ft}}{\text{.}} \cr} $$

Answer: Option C. -> 54 ft
Both the man and pole are near each other and are illuminated by same sun from same direction.
So angle of elevation for sun is same for both.
So ration of object to shadow will be same for all objects. (Proportionality Rule)
$$\frac{{{\text{Object}}\,{\text{height}}}}{{{\text{Shadow}}\,{\text{length}}}} = \frac{6}{4} = \frac{{\text{H}}}{{36}}$$
∴ H = 54 ft = Height of pole