Answer: Option C. -> 546 m
Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 200 m, ∠BAD = 30°, ∠BCD = 45°
$$\eqalign{
& \tan {30^ \circ } = \frac{{BD}}{{BA}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{200}}{{BA}} \cr
& \Rightarrow BA = 200\sqrt 3 \cr
& \tan {45^ \circ } = \frac{{BD}}{{BC}} \cr
& \Rightarrow 1 = \frac{{200}}{{BC}} \cr
& \Rightarrow BC = 200 \cr} $$
Distance between the two ships
$$\eqalign{
& = AC = BA + BC \cr
& = 200\sqrt 3 + 200 \cr
& = 200(\sqrt 3 + 1) \cr
& = 200\left( {1.73 + 1} \right) \cr
& = 200 \times 2.73 \cr
& = 546{\text{ }}m \cr} $$

Answer: Option D. -> 24.8 m
Consider the diagram shown above where PR represents the ladder and RQ represents the wall.
$$\eqalign{
& \cos {60^ \circ } = \frac{{PQ}}{{PR}} \cr
& \frac{1}{2} = \frac{{12.4}}{{PR}} \cr
& PR = 2 \times 12.4 \cr
& \,\,\,\,\,\,\,\,\,\, = 24.8\,m \cr} $$

Answer: Option B. -> 2.16 meter/sec
Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, ∠BCA = 60°
$$\eqalign{
& \tan {60^ \circ } = \frac{{BA}}{{CA}} \cr
& \sqrt 3 = \frac{{BA}}{{150}} \cr
& BA = 150\sqrt 3 \cr} $$
i.e, the distance travelled by the balloon = $$150\sqrt 3 $$   meters
time taken = 2 min = 2 × 60 = 120 seconds
$$\eqalign{
& {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr
& = \frac{{150\sqrt 3 }}{{120}} = 1.25\sqrt 3 \cr} $$
= 1.25 × 1.73 = 2.16 meter/second

Answer: Option B. -> 60°
Let RQ be the pole and PQ be the shadow
Given that RQ = 18 m and PQ = $$6\sqrt 3 $$ m
Let the angle of elevation, ∠RPQ = $$\theta $$
$$\eqalign{
& {\text{From the right}}\Delta PQR, \cr
& {\text{tan}}\,\theta = \frac{{RQ}}{{PQ}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{6\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \cr
& \Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {60^ \circ } \cr} $$

Answer: Option D. -> 54.6 m
Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have ∠DAC = 30°, ∠DBC = 45°
Let DC = h
$$\eqalign{
& \tan {30^ \circ } = \frac{{DC}}{{AC}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{AC}} \cr
& \Rightarrow AC = h\sqrt 3 \,.........\,\left( {\text{i}} \right) \cr
& \tan {45^ \circ } = \frac{{DC}}{{BC}} \cr
& \Rightarrow 1 = \frac{h}{{BC}} \cr
& \Rightarrow BC = h\,........\,\left( {{\text{ii}}} \right) \cr
& {\text{We know that,}} \cr
& AB = \left( {AC - BC} \right) \cr
& \Rightarrow 40 = \left( {AC - BC} \right) \cr} $$
$$ \Rightarrow 40 = \left( {h\sqrt 3 - h} \right)$$   [∵ from (1) and (ii)]
$$\eqalign{
& \Rightarrow 40 = h\left( {\sqrt 3 - 1} \right) \cr
& \Rightarrow h = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \cr
& = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr
& = \frac{{40\left( {\sqrt 3 + 1} \right)}}{{\left( {3 - 1} \right)}} \cr
& = \frac{{40\left( {\sqrt 3 + 1} \right)}}{2} \cr
& = 20\left( {\sqrt 3 + 1} \right) \cr
& = 20\left( {1.73 + 1} \right) \cr
& = 20 \times 2.73 \cr
& = 54.6\,m \cr} $$

Answer: Option A. -> $$60\sqrt 5 \,{\text{m}}$$
Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $$\theta $$
From "Angle Bisector Theorem", we have
$$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$
$$ \Rightarrow \frac{1}{9} = \frac{{15}}{{AC}}$$     [∵ BD : DC = 1 : 9 and AB = 15(given)]
$$ \Rightarrow AC = 15 \times 9\,{\text{m}}\,......\left( {eq:1} \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$
$$CB = \sqrt {A{C^2} - A{B^2}} $$     (∵ Pythagorean theorem)
$$ = \sqrt {{{\left( {15 \times 9} \right)}^2} - {{15}^2}} $$     [∵AC=15 × 9(eq : 1) and AB=15 m(given)]
$$\eqalign{
& = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} \cr
& = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} \cr
& = \sqrt {{{15}^2} \times 80} \cr
& = \sqrt {{{15}^2} \times 16 \times 5} \cr
& = 15 \times 4 \times \sqrt 5 \cr
& = 60\sqrt 5 \,{\text{m}} \cr} $$

Answer: Option A. -> $$24\sqrt 3 \,{\text{km/hr}}$$
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60°
$$\eqalign{
& {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABD, \cr
& \tan {30^ \circ } = \frac{{BD}}{{BA}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{50}}{{BA}} \cr
& \Rightarrow BA = 50\sqrt 3 \cr
& {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CBD, \cr
& \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr
& \Rightarrow \sqrt 3 = \frac{{50}}{{BC}} \cr
& \Rightarrow BC = \frac{{50}}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 }}{3} \cr} $$
Distance between the two persons
$$\eqalign{
& = AC = BA + BC \cr
& = 50\sqrt 3 + \frac{{50\sqrt 3 }}{3} \cr
& = \sqrt 3 \left( {50 + \frac{{50}}{3}} \right) \cr
& = \frac{{200\sqrt 3 }}{3}\,{\text{m}} \cr} $$
i.e., the distance travelled by the car in 10 seconds = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$
$$\eqalign{
& {\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr
& = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}} \cr
& = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}} \cr
& = {\text{24}}\sqrt 3 \,{\text{km/hr}} \cr} $$

Answer: Option C. -> 45°
Consider the diagram shown above where QR represents the tree and PQ represents its shadow
$$\eqalign{
& {\text{We}}\,{\text{have}},\,QR = PQ \cr
& {\text{Let}}\,\angle QPR = \theta \cr} $$
$$\tan \theta = \frac{{QR}}{{PQ}} = 1$$     (since QR = PQ)
$$ \Rightarrow \theta = {45^ \circ }$$
i.e., required angle of elevation = 45°

Answer: Option A. -> 26.28 km/hr
Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m
$$\eqalign{
& \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr
& \Rightarrow 1 = \frac{{AB}}{{100}} \cr
& \Rightarrow AB = 100\,......\left( {eq:1} \right) \cr} $$
$$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$
$$ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{BD}}$$     (∵ Substituted the value of AB from equation 1)
$$ \Rightarrow BD = 100\sqrt 3 $$
$$\eqalign{
& CD = \left( {BD - BC} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\, = \left( {100\sqrt 3 - 100} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\, = 100\left( {\sqrt 3 - 1} \right) \cr} $$
It is given that the distance CD is covered in 10 seconds.
i.e., the distance $$100\left( {\sqrt 3 - 1} \right)$$     is covered in 10 seconds.
$$\eqalign{
& {\text{Required}}\,{\text{speed}} \cr
& = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr
& = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{10}} \cr
& = 10\left( {1.73 - 1} \right) \cr
& = 7.3\,{\text{meter/seconds}} \cr
& = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}} \cr
& = 26.28\,{\text{km/hr}} \cr} $$

Answer: Option A. -> 381 m
Let C and D be the position of the aeroplanes.
Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45°
$$\eqalign{
& {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC, \cr
& \tan {60^ \circ } = \frac{{CB}}{{AB}} \cr
& \sqrt 3 = \frac{{900}}{{AB}} \cr
& AB = \frac{{900}}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\, = \frac{{900 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\, = \frac{{900\sqrt 3 }}{3} \cr
& \,\,\,\,\,\,\,\,\,\, = 300\sqrt 3 \cr
& {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABD, \cr
& \tan {45^ \circ } = \frac{{DB}}{{AB}} \cr
& 1 = \frac{{DB}}{{AB}} \cr
& DB = AB = 300\sqrt 3 \cr
& \cr
& {\text{Required}}\,{\text{height}} \cr
& = CD \cr
& = \left( {CB - DB} \right) \cr
& = \left( {900 - 300\sqrt 3 } \right) \cr
& = \left( {900 - 300 \times 1.73} \right) \cr
& = \left( {900 - 519} \right) \cr
& = 381\,{\text{m}} \cr} $$