Chapter 10, Problem 6RP

To determine

To find : The angular value of $\stackrel{⌢}{AB}$ and $\stackrel{⌢}{AD}$

Answer to Problem 6RP

  $\stackrel{⌢}{AB}=40°$ and $\stackrel{⌢}{AD}=20°$

Explanation of Solution

Given information :

  $\Delta ABC$ is isosceles with base AB i.e. BC = AC, $\angle DAC=70°,\stackrel{⌢}{BC}=160°.$

  

Formula Used:

The angle subtended by an arc of a circle at its centre is twice of the angle it subtends anywhere on the circle's circumference.

Calculation :

  $\begin{array}{l}\frac{1}{2}\angle BOC=\frac{1}{2}\times 160°=80°\\ \Rightarrow \angle BAC=80°\end{array}$

Since, $\Delta ABC$ is isosceles.

  $\begin{array}{l}\angle ABC=\angle BAC=80°\\ \angle BCA=20°(\because \angle ABC+\angle BCA+\angle BAC=180°)\end{array}$

Now, OA = OB = OC (Radius of circle)

From $\Delta OAC$ and $\Delta OBC$

AC = BC ( $\Delta ABC$ is isosceles)

OA = OB (Radius of circle)

OC = OC (common side)

  $\therefore \Delta OAC\cong \Delta OBC$ (SSS Rule)

  $\Rightarrow \frac{1}{2}\angle BCA=\angle BCO=\angle ACO=10°$

Now $\Delta CDA$ is isosceles triangle

  $\angle OAC\ne \angle OCA=10°$

  $\Delta OAB$ is also an isosceles triangle.

  $\angle OAB=\angle OBA=70°[\angle oab+\angle oac=\angle bac]$

  $\Delta BOA$ gives $\angle OAB+\angle OBA+\angle BOA=180°$

  $\begin{array}{l}\Rightarrow 70+70+\angle BOA=180°\\ \Rightarrow \angle BOA=40°\\ \Rightarrow \stackrel{⌢}{AB}=\angle BOA=40°\end{array}$

Again, $\Delta OAB$ is also isosceles triangle.

  $\begin{array}{l}\angle OAD=\angle ODA=80°\\ \angle OAD+\angle ODA+\angle AOD=180°\\ \angle AOD=20°\\ AD=20\end{array}$