Question

Write True or False and justify your answer in each of the following :

If angle between two tangents drawn from a point P to a circle of radius a and centre O is ${60}^0$, then OP $=a\sqrt{3}$.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is B False

$Given-PA\&PBaretangentstothecirclewithcenterOatA\&Brespectively.RadiiOA\&OBare=a\angle APB={60}^{o}Tofindout-Thestatement,OP=a\sqrt{3},istrueorfalse.Justification-PA\&PBaretangentstothecirclewithatA\&Brespectively.\therefore \angle OAP={90}^o=\angle OBP.....\left(i\right)so\Delta OAP\&\Delta OBParerighttriangleswithhypotenuseasOPNowbetween\Delta OAP\&\Delta OBPwehave\angle OAP={90}^o=\angle OBP.....\left(fromi\right)sideOA=sideOBandthehypotenuseOPiscommon.\therefore ByRHStest\Delta OAP\&\Delta OBParecongruent.⟹\angle OPA=\angle OPB=\frac{{60}^o}{2}={30}^o........\left(ii\right)so\angle PAQ={90}^O-{30}^o={60}^o⟹\angle PBA={60}^{o}So\Delta PABisequilateral⟹AB=APAgaininquadrilateralOAPB\angle A=\angle B={90}^o\&\angle P={60}^o\therefore \angle AOB=3{60}^o-({90}^o+{90}^o+{60}^o)={120}^o......\left(iii\right)andSoin\Delta AOB\angle OAQ+\angle OBQ={180}^o-{120}^o={60}^o.ButOA=OB(radiiofthesamecircle.So\Delta AOBisanisoscelesone.\therefore \angle OAQ=\angle OBQ={30}^o\therefore in\Delta APO\angle OPA+\angle PAO+\angle AOP={180}^o(anglesumpropertyoftriangles)⟹{30}^o+{90}^o+\angle AOP={180}^o⟹\angle AOP={60}^o....(iii)Againin\Delta AOQ\angle AQO=\angle OAQ+\angle AOQ+\angle OQA={180}^o(anglesumpropertyoftriangles)⟹{30}^o+{60}^o+\angle OQA={180}^o⟹\angle OQA={90}^o⟹\Delta OQAisarightone.Sobetween\Delta OQA\&\Delta OPAwehavetheanglesequali.etheyaresimilar.\therefore \frac{AQ}{a}=\frac{AP}{OP}⟹\frac{AB}{2a}=\frac{AP}{OP}⟹OP=2a(AB=AP)SotheThestatement,OP=a\sqrt{3},isfalse.Ans-False$