**Question 5**:

In figure above,

*PAQ*is a tangent to the circle at point*A*.*AEC*and*BED*are straight lines. The value of*y*is

*Solution:*∠

*ABD*= ∠*ACD =*40^{o}∠

*ACB*= ∠*PAB =*60^{o}*y*= 180

^{o }– ∠

*ACB*– ∠

*CBD*– ∠

*ABD*

*y*= 180

^{o }– 60

^{o}– 25

^{o}– 40

^{o}=

**55**

^{o}**Question 6**:

In figure above,

*KPL*is a tangent to the circle*PQRS*at point*P*. The value of*x*is

Solution:Solution:

∠

*PQS*= ∠*SPL*= 55^{o}∠

*SPQ*= 180^{o }– 30^{o}– 55^{o}= 95^{o}In cyclic quadrilateral,

∠

*SPQ*+ ∠*SRQ*= 180^{o}95

^{o}+*x*^{o}= 180^{o}

*x***= 85**

^{o}**Question 7**:

In figure above,

*APB*is a tangent to the circle*PQR*at point*P.**QRB*is a straight line. The value of*x*is

*Solution:*∠

*PQR*= ∠*RPB*= 45^{o}∠

*QPR*= (180^{o }– 45^{o}) ÷ 2 = 67.5^{o}∠

*PQR*+ ∠*BPQ + x*^{o}= 180^{o}45

^{o}+ (67.5^{o}*+ 45*^{o}) +*x*^{o}= 180^{o }

*x***= 22.5**

^{o}**Question 8**:

The figure above shows two circles with respective centres

*O*and*V*.*AB*is a common tangent to the circles.*OPRV*is a straight line. The length, in cm, of*PR*is

Solution:Solution:

**$\begin{array}{l}\mathrm{cos}{86}^{o}=\frac{OM}{OV}\\ 0.070=\frac{1}{OV}\\ OV=\frac{1}{0.070}\\ OV=14.29cm\\ \\ \therefore PR=14.29-5-4\\ \text{}=5.29cm\end{array}$**