 |
$\begin{align}(i)x=&\frac{100}{2}\\[5pt]
=&\underline{\underline{50^0}}\\
\end{align}$ | $\begin{align}(ii)x=&2\times 30^0\\[5pt]
=&\underline{\underline{60^0}}\\
\end{align}$ | $\begin{align}(iii)2x&=50\\[5pt]
x&=\frac{50}{2}\\[5pt]
=&\underline{\underline{25^0}}\\
\end{align}$ |
$\begin{align}(iv)x=&\frac{310}{2}\\[5pt]
=&\underline{\underline{155^0}}\\
\end{align}$ | $\begin{align}(v)x=&360-160=200^0\\[5pt]
x=&\frac{200}{2}\\[5pt]
=&\underline{\underline{100^0}}\\
\end{align}$ | $\begin{align}(vi)x+x=&70^0\\[5pt]
2x=&70^0\\[5pt]
x=&\frac{70}{2}\\[5pt]
=&\underline{\underline{35^0}}\\
\end{align}$ |
 |
$\begin{align}(i)y=&\frac{100}{2}\\[5pt]
=&\underline{\underline{50^0}}\\
&\text{(Angle subtended at the centre is }\\
&\text{twice as angle subtended }\\
&\text{on the remaining part of circle)}\\
\end{align}$ | $\begin{align}(ii)O\hat AC=&y\text{ }\text{(OAC is an isosceles triangle)}\\[5pt]
y+y=&90^0\\[5pt]
&\underline{\underline{y=45^0}}\\
&\text{(Exterior angle is sum of }\\
&\text{opposite interior angles)}\\
\end{align}$ | $\begin{align}(iii)B\hat AC=&180-(70+60)\\[5pt]
=&180-130\\[5pt]
&\underline{\underline{=50^0}}\\
&\text{(Sum of angles}\\
&\text{in a triangle)}\\
y=&50^0\times 2\\[5pt]
&\underline{\underline{y=100^0}\text {(theorem 1)}}\\
\end{align}$ |
$\begin{align}(iv)C\hat BO=&20^0\text{(Isosceles triangle)}\\[5pt]
A\hat BO=&30^0\text{(Isosceles triangle)}\\[5pt]
A\hat BC=&30+20=50^0\\[5pt]
y=&2\times 50\\[5pt]
&\underline{\underline{y=100^0}\text {(theorem 1)}}\\
\end{align}$ | $\begin{align}(v)P\hat OR=&110\times 2^0\text{(Reflex)}\\[5pt]
&\underline{\underline{=220^0}}\\
y=&360-220\\[5pt]
&\underline{\underline{y=140^0}\text {(Angle at a point)}}\\
\end{align}$ | $\begin{align}(vi)30+y=&\frac{110}{2}\\[5pt]
30+y=&55^0\\[5pt]
y=&55-30\\[5pt]
&\underline{\underline{y=25^0}}\\
\end{align}$ |
$\begin{align}(vii)B\hat OC=&70\times 2=140\text {(theorem)}\\[5pt]
2x+140=&180\text {(Sum of triangle)}\\[5pt]
x=&\frac{40}{2}\\[5pt]
x=&20^0\\[5pt]
180^0=&70+y+20+y+20\\[5pt]
2y=&180^0-110^0\\[5pt]
y=&70^0\\[5pt]
&\underline{\underline{y=35^0}}\\
\end{align}$ | $\begin{align}(viii)Q\hat OR=&180-100\\[5pt]
=&80^0\\[5pt]
y=&\frac{80}{2}\\[5pt]
&\underline{\underline{y=40^0}}\\
\end{align}$ | $\begin{align}(ix)B\hat AC=&\frac{130}{2}\\[5pt]
=&65^0\text{(Theorem 1)}\\[5pt]
y+65=&180\text{(Straight angle)}\\[5pt]
y=&180-65\\[5pt]
&\underline{\underline{y=115^0}}\\
\end{align}$ |
By studying this lesson you will be able to
Identify and apply the theorems related to angles in a circle.