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Question

If PA and PB are tangents from an outside point P. such that PA =10 cm and $$\angle APB = 60$$. Find the length of chord AB.
971268_455de95b1b1545fabb9a0283e0a217a9.png


Solution


Consider a circle C with center O.

We have PA and PB are tangents of the circle, PA=10cm and 
$$\angle APB=60^{o}$$

Join OP,

Such that,

In $$\triangle PAC$$ and $$\triangle PBC$$ we have,

$$PA=PB$$   [tangent of the circle fro the outer point p is equal]

$$\angle PAC= \angle PBC$$   [angle made by the external tangent on a 

circle is equal]

$$PC=CP$$  [common]

so,

$$\triangle PAC$$ $$\cong$$ $$\triangle PBC$$  [By SAS criteria]

so,

$$AC=BC$$ ........(i)

$$\angle ACP= \angle BCP$$ ......(ii)

since

$$\angle APB= \angle APC+ \angle BPC$$

so,

$$\angle APC= \dfrac {1} {2} \times 60^0=30^0$$ 

$$\angle APC=30^0$$

$$\angle ACP+ \angle BCP=180^0$$

from equation 2 we get

$$\angle ACP=\dfrac {1} {2} \times180^0=90^0$$

Thus in Right $$\triangle$$ ACP

$$sin$$$$30^0=\dfrac{AC}{AP}$$

$$\dfrac{1}{2}=\dfrac{AC}{10}$$cm

$$AC=5$$cm

Since 

$$AC=BC$$

so,

$$AB= AC+BC$$

     $$=5$$cm$$+5$$cm

     $$=10$$cm


951835_971268_ans_57e83154b811456d96719f884086f59e.png

Mathematics

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