Question

if the angles of elevation of the top of a vertical tower from two points A and B on the ground are respectively 30 and 60.Then the ratio of the distance of A and B from the upper end of the tower is.... a) √3:1, b) 1:√3, c) (√3+1):1, d) 1:(√3+1) ​

Answer 1

Answer:

option b)

Step-by-step explanation:

tan30=h/x+y

√3h=x+y

h=x+y/√3

then, tan60=h/y

h=√3y

now equating the values of h,

x+y/√3=√3y

2y=x we get

then apply pythagorous therom,

then we get ratio as 1:√3

Answer 2

a) √3 : 1 is the ratio of the distance of A and B from the upper end of the tower

Step-by-step explanation:

Given: The angles of elevation from point A = 30°

The angles of elevation from point B = 60°

To Find: The ratio of the distance of A and B from the upper end of the tower

Solution:

• Finding the ratio of the distance of A and B from the upper end of the tower

Considering the triangle OAC such that

$sin 30^{o} = \frac{OC}{AC}$

$\Rightarrow \frac{OC}{AC} = \frac{1}{2} \ \cdots \cdots (1)$

Similarly, considering the triangle ObC such that

$sin 60^{o} = \frac{OC}{BC}$

$\Rightarrow \frac{OC}{BC} = \frac{\sqrt{3}}{2} \ \cdots \cdots (2)$

Divide (2) by (1), we get,

$\frac{AC}{BC} = \frac{sin 60^{o}}{sin 30^{o}} = \sqrt{3} : 1$

Hence, √3 : 1 is the ratio of the distance of A and B from the upper end of the tower