Question

In fig. 1, PT and PT' are tangents to the circle with center O. If $\angle TR{T}^{\prime }={70}^{\circ }$, then x equals

• A. ${30}^{\circ }$
• B. ${35}^{\circ }$
• C. ${40}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

Answer: (C)

${40}^{\circ }$

$\angle TO{T}^{\prime }=2\angle TR{T}^{\prime }$ (Angle subtended by the same arc is twice at the centre than on the circumference of a circle)

So, $\angle TO{T}^{\prime }=2\angle TR{T}^{\prime }={140}^{\circ }$

And we know $\angle PTO$ and $\angle P{T}^{\prime }O$ are ${90}^{\circ }$

In quadrilateral OTPT'

$\angle OTP+\angle TP{T}^{\prime }+\angle P{T}^{\prime }O+\angle TO{T}^{\prime }={360}^{\circ }$

${90}^{\circ }+\angle TP{T}^{\prime }+{90}^{\circ }+{140}^{\circ }={360}^{\circ }$

therefore $x={40}^{\circ }$

Option C is the answer.