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3 years ago

What is the length of RS.

Mathematics

1 answer:

Natali [406]3 years ago

5 0

Consider, in ΔRPQ,

RP = R (Radius of larger circle)

PQ = r (radius of smaller circle)

We have to find, RQ, by Pythagoras theorem,

RP² = PQ²+RQ²

R² = r²+RQ²

RQ² = R²-r²

RQ = √(R²-r²

Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ

RS = 2√(R²-r²)

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______________________________________________________

TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-

For any right angled triangle with one angle α ,

$\cos (90 - \alpha ) = \sin \alpha$ or $\sin(90 - \alpha ) = \cos\alpha$

$cosec \: (90 - \alpha ) = \sec\alpha$ or $\sec(90 - \alpha ) = cosec\:\alpha$

SOME GENERAL TRIGNOMETRIC FORMULAS :-

$\sin \alpha = \frac{1}{cosec \: \alpha }$ or $cosec \: \alpha = \frac{1}{\sin \alpha }$

$\cos \alpha = \frac{1}{\sec \alpha }$ or $\sec \alpha = \frac{1}{\cos \alpha }$

______________________________________________________

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

$\cos (90 - \alpha ) \times cosec(90 - \alpha )$

$=> sin\alpha \times \sec\alpha$

$=> \sin\alpha \times \frac{1}{\cos\alpha }$

$=> \frac{\sin\alpha }{\cos\alpha }$

$=> \tan\alpha$ = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

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