Solution:

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°

Solution:

P : Q
1 : √3

Solution:

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
$$\eqalign{ = AC = BA + BC \cr = 200\sqrt 3 + 200 \cr = 200(\sqrt 3 + 1) \cr = 200\left( {1.73 + 1} \right) \cr = 200 \times 2.73 \cr = 546{\text{ }}m \cr} $$

Solution:

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 = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$
$$\eqalign{ {\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}} \cr = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}} \cr = {\text{24}}\sqrt 3 \,{\text{km/hr}} \cr} $$

Solution:

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)]
$$\eqalign{ = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} \cr = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} \cr = \sqrt {{{15}^2} \times 80} \cr = \sqrt {{{15}^2} \times 16 \times 5} \cr = 15 \times 4 \times \sqrt 5 \cr = 60\sqrt 5 \,{\text{m}} \cr} $$

Solution:

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

Solution:
Let AB be the tower.

Then, ∠APB = 30° and AB = 100m
$$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
⇒ AP = AB × √3 = 100 × √3
⇒ AP = 100 × 1.73 = 173m