Question

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

Answer 1

Answer:

Height =5 ×1.73 m

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Answer 2

The height is $5\sqrt{3}$

Given:

Distance apart on a line inclined at an angle of 15° = 10cm

Angles of elevation of the object from P and Q are 30° and 60°

To Find:

The height

Solution:

Let AB = h

PQE is the 15° inclined surface.

PQ = 10cm

∠BPE = 15°, ∠EPA = 30°, ∠EQA = 60°

In ΔPQA we have ∠P = 30°, ∠Q = 120° and ∠A = 30°

Using sine rule:

$\implies \frac{AP}{sin 120} = \frac{PQ}{sin 30} \\\\\implies \frac{AP}{\sqrt{3}/2 } = \frac{PQ}{1/2} \\\\AP = \sqrt{3} \times 10\\\\\implies AP = 10\sqrt{3}$

Now, consider ΔPBA,

$\implies sin 30 = \frac{AB}{AP} \\\\\implies \frac{1}{2} = \frac{h}{10\sqrt{3} } \\\\\implies h = \frac{10\sqrt{3}}{2} \\\\\implies h = 5\sqrt{3}$

Therefore, the height is $5\sqrt{3}$

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