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<h3><u>Given</u><u> </u><u>:</u><u>-</u></h3>
- PQ = 8cm
- Radius = 5cm
- Two Tangents = P & Q.
- Join OT.
Here, ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.
[∵ TP=TQ = Tangents from T upon the circle]
⠀⠀⠀⠀⠀⠀⠀⠀∴ OT⊥PQ
- So, PR = RQ = 4cm.
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By Applying Pythagoras Theorem in ∆OPR :
OR = √OP² - PR²
OR = √5² - 4²
OR = 3cm
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Now,
∠TPR + ∠RPO = 90° (∵TPO=90°)
∠TPR + ∠PTR (∵TRP=90°)
<u></u><u>∴ ∠RPO = ∠PTR</u>
⠀⠀
<u>∴ Right triangle TRP is similar to the right </u><u>triangle</u><u> </u><u>PR</u><u>O</u><u>.</u> [By A-A Rule of similar triangles]
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<h2>-MissAbhi</h2>
The answer is x=6/7.
