Question

If two towers of height h1 and h2 subtends angles of 30 and 60 respectively at the midpoints of line joining their feets.find h1:h2

Answer 1

Answer:

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Answer 2

SOLUTION:-

Given:

Two towers of height h1 & h2.

Subtends angle 30° & 60°

Let O be the mid-point of the line joining their feet.

$= > BC = \frac{1}{2} OB = \frac{1}{2} OC.............(1)$

Now,

In ∆AOB,

$tan 60 \degree = \frac{Perpendicular}{Base} = \frac{AB}{OB} \\ \\ = > tan60 \degree = \frac{h1}{2BC} \: \: \: \: \: [from \: (1)] \\ \\ = > 2 \sqrt{3BC} = h1...............(2)$

And,

In ∆DOC,

$tan30 \degree = \frac{Perpendicular}{Base} = \frac{DC}{OC} \\ \\ = > tan30 \degree = \frac{h2}{2BC} \: \: \: \: \: [from \: (1)] \\ \\ = > \frac{2BC}{ \sqrt{3} } = h2.............(3)$

So,

Mid-points of line joining their feets:

h1:h2

$= > h1 \ratio \: h2 = 2 \sqrt{3BC} \ratio \: \frac{2BC}{ \sqrt{3} } \\ \\ = > h1 \ratio \: h2 = \frac{2 \sqrt{3BC} }{ \frac{2BC}{ \sqrt{3} } } = \frac{2 \sqrt{3BC \times \sqrt{3} } }{2BC} \\ \\ = > \frac{ \sqrt{3 } \times \sqrt{3} }{1} \\ \\ = > \frac{3}{1} \\ \\ = > h1 \ratio h2 = 3 \ratio 1$

Hope it helps ☺️