1. The chord of a circle is 10 cm in length and at a distance of $$\frac{10}{\sqrt3}$$cm from the centre. Calculate the angle subtended by the chord at the centre of a circle.
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By: anil on 05 May 2019 03.00 pm
Given : AB = 10 cm and OM = $$frac{10}{sqrt3}$$ cm To find : $$angle$$ AOB = $$ heta$$ = ? Solution : Perpendicular from the centre to the chord bisects the chord, => MB = $$frac{10}{2}=5$$ cm Also, $$angle$$ AOB = $$2angle$$ MOB => $$angle$$ MOB = $$frac{ heta}{2}$$ ------------(i) Now, in $$ riangle$$ OMB => $$tan(angle MOB)=frac{MB}{OM}$$ => $$tan(frac{ heta}{2})=5divfrac{10}{sqrt3}$$ => $$tan(frac{ heta}{2})=5 imesfrac{sqrt3}{10}$$ => $$tan(frac{ heta}{2})=frac{sqrt3}{2}$$ => $$frac{ heta}{2}=tan^{-1}(frac{sqrt3}{2})$$ => $$ heta=2 an^{-1}(frac{sqrt{3}}{2})$$
=> Ans - (A)
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=> Ans - (A)