Question

5. If PT is a tangent to a circle with centre o
and PQ is a chord of the circle such that
OPT = 70°, then find the measure of angle POQ.
​

Answer 1

Explanation:

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Answer 2

Diagram :-

$\setlength{\unitlength}{2mm}\begin{picture}(0,0)\thicklines\put(3,0){\circle{10}}\put(2.9,3.5){\vector(-3,0){1.5cm}}\put(3,0){\circle*{0.3}}\put(2.89,0){\line(0,3){0.7cm}}\put(-3,3.487){\circle*{0.3}}\put(-3.5,4.2){\sf \footnotesize T}\multiput(3,3.5)(1,0){8}{\line(3,0){1mm}}\put(3,2.5){\line(3,0){1.5mm}}\put(3.7,3.5){\line(0,-3){1.62mm}}\put(3.2,-1){\sf\footnotesize O}\put(3,0){\line(-3,-2){5.6mm}}\put(3,3.5){\line(-3,-5){6mm}} \put(3,4){\sf\footnotesize P}\put(-1,-3){\sf\footnotesize Q}\qbezier(2,3.5)(1.5,2)(2.4,2.9)\put(0.4,1.5){\sf\tiny 70^\circ }\end{picture}$

Solution :-

Here ,

• OP $\perp$ Tangent . So , $\angle$ OPT = 90°
• $\angle$QPT = 70°
• OP = OQ [°.° Radii are always equal ]

$\boldsymbol\angle$OPT = 90°

• $\boldsymbol\angle$ OPT = $\boldsymbol\angle$ QPT + $\boldsymbol\angle$OPQ

• $\boldsymbol\angle$ OPQ = 90° - 70°

• $\boldsymbol\angle$ OPQ = 20°

Since the radii are equal . Here we have chord PQ . By observing the given figure ∆OPQ is a an isosceles ∆.

Now , as we know that equal sides opposite angles are always equal in isosceles triangle .

So ,

∠OPQ = ∠POQ

Now , using angle sum property .

⇒ 20° + 20° + ∠POQ = 180°

⇒ 40° + ∠POQ = 180°

⇒ ∠POQ = 180° - 40°

⇒∠POQ = 140°

Hence , ∠POQ = 140°