Question

The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the center of the line joining their feet, then find the ratio of x and y.
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Answer 1

✰ Given :

• The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the center of the line joining their feet.

✰ To Find :

• The ratio of x and y.

✯ Required Solution :

❍ Kindly, see the attachment .

❏ Let, x and y be AB and CD, respectively.

❒ Let, the centre of the line joining of the feet of the two towers ( BD ) be E.

$\\ \qquad \qquad {\underbrace{ \sf{❍ \:\:In \: \Delta}\bf \: ABE \: \bf,}}$

$\bf{✧\:\:Here, }\: \sf\frac{AB}{BE} \: ➠ \: \bf \tan30 {}^{ \small\circ} .$

$: \implies \sf \frac{x}{BE} \: ➠ \: \frac{1}{ \sqrt{3} } \\\\\ : \implies \sf BE \: ➠ \: \sqrt{3x} \: \: \cdots \cdot \bf(1)$

$\qquad \qquad {\underbrace{ \sf{❍ \:\:In \: \Delta}\bf \: CDE \: \bf,}}$

$\bf{✧\:\:Here, }\: \sf\frac{CD}{DE} \: ➠ \: \bf \tan60 {}^{ \small\circ} .$

$: \implies \sf\frac{y}{DE} \: ➠ \: \sqrt{3} \\\\\ : \implies \sf DE \: ➠ \: \frac{y}{ \sqrt{3} } \: \cdots \cdot \bf(2)$

Then, BE ➠ DE $\cdots\cdot \bf(3)$

( We know that, E is mid point of BD )

Then, we get from (1), (2) and (3),

$: \sf \implies \: \sqrt{3x} \: ➠ \: \frac{y}{ \sqrt{3} } \\\\\ : \sf \implies\frac{x}{y} \: ➠ \: \frac{1}{3 }\\\\\ \qquad \quad \underline{ \boxed{\bf : \implies \: 1 : 3}} \: \: ✰$

.°. Hence, the ratio of $x$ and $y$ is 1:3.

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✰ Hope it helps u :)) ✰

Answer 2

Question:-

The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the center of the line joining their feet, then find the ratio of x and y.

Required Answer:-

Given:-

$\\ \sf{ \rightarrow{Height \: of \: the \: towrs \: x \: and \: y}}$

$\sf{ \rightarrow{The \: tops \: subtend \: angles \: 30 \degree \: and \: 60 \degree}}$

To Find:-

$\\ \sf{ \rightarrow{Ratio \: of \: x \: and \: y}}$

♡Solution:-

Let,

AB = x and BC = y

Now, let's assume that E is the center of BD joining the feet of two towers.

We Know that:-

$\\ \sf{ \bold{ \tan(θ) = \frac{Perpendicular}{Base} }}$

$\\ \sf{In} \: \bold{ \triangle{ABE}}$

• AB is the Perpendicular
• BE is the base

$\\ \sf{\therefore{\frac{AB}{BE} = \tan(30 \degree) }}$

$\\ \sf{ \implies{ \frac{x}{BE} = \frac{1}{ \sqrt{3} } }}$

$\\ \sf{ \implies{ \blue{ BE = \sqrt{3} x}}} - - - - - - - - - (1)$

Again,

$\\ \sf{in} \: \bold{ \triangle{CDE}}$

• CD is the perpendicular
• DE is the base

$\\ \sf{ \therefore{ \frac{CD}{DE} = \tan(60 \degree) }}$

$\\ \sf{ \implies{ \frac{y}{DE} = \sqrt{3} }}$

$\\ \sf{ \implies{ \blue{DE = \frac{y}{ \sqrt{3} } }} }- - - - - - - - (2)$

Now,

As, E is the center of BD, we can write:-

$\\ \sf{ \blue{BE = DE}} - - - - - - - - (3)$

From, equation (1) , (2) and (3) we find,

$\\ \sf{ \bold{ \sqrt{3} x = \frac{y}{ \sqrt{3} } }}$

$\\ \sf{ \implies{ \frac{ \sqrt{3}x }{ \sqrt{3} } = \frac{y}{ (\sqrt{3} )( \sqrt{3}) } \: \: \: \: \boxed{ \rm{dividing \: by \: \sqrt{3} }}}}$

$\\ \sf{ \implies{x = \frac{y}{ 3 } }}$

$\\ \sf{ \implies{ \frac{x}{y} = \frac{1}{3} }}$

$\\ \sf{ \therefore{ \bold{x \ratio y }= \red{1 :3}}}$

$\\ \underline{\bold{ \boxed{ \rm{Hence \: ratio \: of \: x \: and \: y = \red{ \bold{1 :3}}}}}}$

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Note :- Diagram refers to the attachment!