Directions:

Two pillars P₁ and P₂ of equal heights stand on either side of a road which is 100 m wide. At a point on the road between the pillars, the angles of elevation of the tops of the pillars P₁ and P₂ are 60° and 30° respectively.

Based on this information, answer the following questions:

30. The height of each pillar is:
(a) 25 m
(b) 36 m
(c) $25\sqrt{3}$ m
(d) $36\sqrt{3}$ m

31. The location of the point of observation is:
(a) 25 m from P₁
(b) 25 m from P₂
(c) $25\sqrt{3}$ m from P₁
(d) $25\sqrt{3}$ m from P₂

32. If a hook is fixed at the point of observation and strings are tied from the hook to the tops of both the towers, then the total length of string required is:
(a) 43.3 m
(b) 86.6 m
(c) 68.3 m
(d) 136.6 m

33. If a flagstaff is to be erected atop pillar P₂ such that the angle of elevation of its top from the point of observation is 45°, then the height of the flagstaff must be:
(a) $25\sqrt{3}$ m
(b) 50 m
(c) $25\sqrt{3}(\sqrt{3}-1)$ m
(d) $25(2-\sqrt{3})$ m

30. Let AP₁ and BP₂ be two pillars of height h.

Let the point of observation be O at distance x from P₁.

In triangle OAP₁,

$\Rightarrow \tan 60^{\circ} = \dfrac{h}{x} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{x} \\[1em] \Rightarrow x = \dfrac{h}{\sqrt3}.$

In triangle OBP₂,

$\Rightarrow \tan (30^{\circ}) = \dfrac{h}{100 - x} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h}{100 - x} \\[1em] \Rightarrow 100 - x = h\sqrt3 \\[1em] \Rightarrow 100 - \dfrac{h}{\sqrt{3}} = h\sqrt{3} \\[1em] \Rightarrow 100 = h\sqrt{3} + \dfrac{h}{\sqrt{3}} \\[1em] \Rightarrow 100 = h \Big( \frac{3 + 1}{\sqrt{3}} \Big) = \dfrac{4h}{\sqrt{3}} \\[1em] \Rightarrow h = \dfrac{100\sqrt{3}}{4} = 25\sqrt{3} \text{ m}.$

Hence, option (c) is the correct option.

31. In triangle OAP₁,

$\Rightarrow \tan 60^{\circ} = \dfrac{h}{x} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{x} \\[1em] \Rightarrow x = \dfrac{h}{\sqrt3} \\[1em] \Rightarrow x = \dfrac{25\sqrt{3}}{\sqrt{3}} = 25 \text{ m}$

Hence, option (a) is the correct option.

32. Let the length of strings be L₁ and L₂.

In triangle OAP₁,

$\Rightarrow \sin 60^{\circ} = \dfrac{h}{L$1} \\[1em] \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{25\sqrt3}{L1} \\[1em] \Rightarrow L_1 = 50 \text{ m}.

In triangle OBP₂,

$\Rightarrow \sin (30^{\circ}) = \dfrac{h}{L$2} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{25\sqrt3}{L2} \\[1em] \Rightarrow L_2 = 50\sqrt3 = 86.6 \text{ m.}

Total length = 50 + 86.6 = 136.6 m

Hence, option (d) is the correct option.

33. Let H be the height of flagstaff added to P₂

Total distance - Distance to P₁ = Distance to P₂

Distance to P₂ = 100 - 25 = 75 m.

Therefore,

$\Rightarrow \tan (45^{\circ}) = \dfrac{H + 25\sqrt3}{75} \\[1em] \Rightarrow 1 = \dfrac{H + 25\sqrt3}{75} \\[1em] \Rightarrow 75 = 25\sqrt3 + H \\[1em] \Rightarrow H = 75 - 25\sqrt3 \\[1em] \Rightarrow H = 25\sqrt{3}(\sqrt{3} - 1) \text{ m.}$

Hence, option (c) is the correct option.