**3.2.1 Circles II, PT3 Practice**

**Question 1:**

*KLM*,

*LPN*and

*MPO*are straight lines.

Find the value of

*x*and of*y*.(b) In the diagram below,

*O*is the centre of the circle. Find the value of

(i)

*x*(ii)*y*

*Solution:***(a)**

$\begin{array}{l}\angle LOM=\angle LNM={y}^{o}={44}^{o}\\ \therefore y=44\\ \angle NMO=\angle NLO={41}^{o}\\ {x}^{o}={180}^{o}-{80}^{o}-{41}^{o}\\ \text{}={59}^{o}\end{array}$

(b)(i)

(b)(i)

2

*x*= 200^{o}

*x***= 100**

^{o}

(b)(ii)

(b)(ii)

*x*+

*y*= 180

^{o}

100

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

*y***= 80**

^{o }**Question 2:**

*AOC*and

*BCD*are straight lines with centre

*O*.

(a) In Diagram below,

Find the value of

*y*.*ABCD*is a circle with centre

*O*and

*ADE*is a straight line.

(b) In diagram below,

Find the value of

*y*.

*Solution:***(a)**

$\begin{array}{l}\angle AOB={45}^{o}\times 2={90}^{o}\\ {y}^{o}={180}^{o}-{90}^{o}={90}^{o}\end{array}$

(b)

(b)

$\begin{array}{l}\angle ABD=\frac{1}{2}\times {100}^{o}={50}^{o}\\ \angle CDE=\angle ABC={85}^{o}\\ {y}^{o}+{50}^{o}={85}^{o}\\ \text{}{y}^{o}={35}^{o}\\ \text{}y=35\end{array}$

**Question 3:**

(a)

*ABCD*is a circle.*AEC*and*BED*are straight lines.**In diagram below,**Find the value of

*y*.*KLMN*with centre

*O*.

**(b) Diagram below shows a circle**

Find the value of

*x*.

Solution:Solution:

(a)

$\begin{array}{l}\angle ABE=\angle ACD={45}^{o}\\ {y}^{o}={180}^{o}-{45}^{o}-{55}^{o}\\ {y}^{o}={80}^{o}\\ y=80\end{array}$

(b)

$\begin{array}{l}\angle OML=\angle OLM={42}^{o}\\ \left({x}^{o}+{42}^{o}\right)+\angle KNM={180}^{o}\\ {x}^{o}+{42}^{o}+{110}^{o}={180}^{o}\\ {x}^{o}+{152}^{o}={180}^{o}\\ {x}^{o}={180}^{o}-{152}^{o}\\ {x}^{o}={28}^{o}\\ x=28\end{array}$

**Question 4:**

Diagram below shows a semicircle

*ABCD*with centre*O.*

It is given that

*AB*=*CD*.Find the value of

*y*.

*Solution:**y*

^{o }+ 32

^{o}+ 32

^{o}+ 90

^{o}= 180

^{o}

*y*

^{o }= 180

^{o}– 154

^{o}

*y*

^{o }= 26

^{o}

*y***= 26**

**Question 5:**

In diagram below,

*ABC*and

*DEF*are straight lines.

Find the value of

*x*and of

*y*.

*Solution:*$\begin{array}{l}x+38=93\\ \text{}x=93-38\\ \text{}x=55\\ \\ \angle DBG={38}^{o}\text{}\left(GB=GD\right)\\ \text{}{y}^{o}=\angle DBG\\ \text{}\therefore y=38\end{array}$