Question

Let an object be placed at some height h cm and let P and Q be two points of observation which are
at a distance 10 cm apart on a line inclined at angle 15° to the horizontal. If the angles of elevation of
the object from P and Q are 30°and 60° respectively then find h.​

Answer 1

Consodering both the triangles,

In first triangle,

tan 30 = h/x+10

And in other triangle,

tan 60 = h/x

x = h/tan 60

Now put in equation 1,

tan 30 = h/(h/tan 60)+10

0.57=h/0.57h+10

0.325h + 5.7 = h

5.7 = 0.675h

h = 5.7/0.675

h = 8.44

Answer 2

Given :-

• An object be placed at some height h cm
• P and Q be two points of observation which are at a distance 10 cm apart on a line inclined at angle 15° to the horizontal
• The angles of elevation of the object from P and Q are 30°and 60° respectively.

To Find :-

• What is it's height ?

Solution :-

Given,

• AB = h
• PQ = 10cm

Let,

Angle of elevation from P and Q are,

• ∠BPA = 30°
• ∟A = 90°

From ∆PQB, (Using sine rule)

$\qquad \displaystyle \sf \frac{PQ }{Sin \: 30°} = \frac{ PB} {Sin \: 135°}$

$\qquad \displaystyle \sf \frac{10}{ \displaystyle \frac{1}{ 2 } } = \frac{ PB} { \displaystyle \frac{1}{ \sqrt{2} } }$

$\qquad \displaystyle \sf 10 \times 2= PB \sqrt{2}$

$\qquad \displaystyle \sf 10 \times \sqrt{2} \times \cancel{ \sqrt{2} }= PB \: \cancel{\sqrt{2} }$

$\qquad \displaystyle \sf PB = 10 \sqrt{2}$

From ∆PAB,

$\qquad \displaystyle \sf Sin \: 30° = \frac{ AB} {PB} = \frac{h}{10 \sqrt{2} }$

$\qquad \displaystyle \sf \frac{1} {2} = \frac{h}{10 \sqrt{2} }$

$\qquad \displaystyle \underline{\boxed{\bf h = 5\sqrt{2} }}$

Hence,

∴ The height is 5√2 .

$\:$