Solveeit Logo

Question

Question: A man on the top of a cliff 100 m high, observe the angles of depression of two points on the opposi...

A man on the top of a cliff 100 m high, observe the angles of depression of two points on the opposite sides of the cliff as 30{{30}^{\circ }} and 60{{60}^{\circ }} respectively. Then, the distance between the two points is equal to
(a)4003m400\sqrt{3}m
(b)4003m\dfrac{400}{\sqrt{3}}m
(c)1003m\dfrac{100}{\sqrt{3}}m
(d)2003200\sqrt{3}

Explanation

Solution

Hint: For solving this problem, we consider two triangles ABD and ADC having respective bases as a and b in metres. Now, by applying the trigonometric ratio of tanθ\tan \theta in both the triangles, we obtain the respective bases individually as height is already given in the problem statement. By adding both the bases of the triangle, we get the final answer.

Complete step-by-step answer:

Let, the base of triangle ABD be ‘a’ meters and base of triangle ADC be ‘b’ meters.
According to the problem statement, the height of the cliff is 100m, ABD=60\angle ABD={{60}^{\circ }} and ACD=30\angle ACD={{30}^{\circ }}.
One useful trigonometric ratio involved in this problem is tan which can be expressed as:
tanθ=perpendicularbase\tan \theta =\dfrac{perpendicular}{base}
In ΔABD\Delta ABD, applying tan60{{60}^{\circ }}, we get
tan60=ADBD\tan {{60}^{\circ }}=\dfrac{AD}{BD}
As we know that the value of tan60=3\tan {{60}^{\circ }}=\sqrt{3}, AD =100m and BD = a meter. Therefore, to evaluate a we put the values in the above ratio.
3=100a a=1003m \begin{aligned} & \sqrt{3}=\dfrac{100}{a} \\\ & a=\dfrac{100}{\sqrt{3}}m \\\ \end{aligned}
The length of BD is 1003m\dfrac{100}{\sqrt{3}}m.
In ΔADC\Delta ADC, applying tan30\tan {{30}^{\circ }}, we get
tan30=ADDC\tan {{30}^{\circ }}=\dfrac{AD}{DC}
As we know that the value of tan30=13{{30}^{\circ }}=\dfrac{1}{\sqrt{3}}, AD = 100 m and DC = b meter.
13=100b b=1003m \begin{aligned} & \dfrac{1}{\sqrt{3}}=\dfrac{100}{b} \\\ & b=100\sqrt{3}m \\\ \end{aligned}
The value of DC is 1003m100\sqrt{3}m,
Adding both the value of BD and DC with each other, we get
a+b=1003+1003 a+b=4003m \begin{aligned} & a+b=\dfrac{100}{\sqrt{3}}+100\sqrt{3} \\\ & a+b=\dfrac{400}{\sqrt{3}}m \\\ \end{aligned}
Hence, the distance between two points is 4003m\dfrac{400}{\sqrt{3}}m.
Therefore, option (b) is correct.
Note: The key concept involved in solving this problem is the knowledge of trigonometric ratio in the form of triangles. Once we know the value of angle subtended, we can easily find the ratio between the length by using tan operation.