Question

can anybody tell me the example no.8 of chapter 6th of maths class 9 ncert
it's emergency
tell me plz tommorow is my exam !!
also send me a pic of example 8th

Answer 1

Your answer :-

$\textbf {Given => BO bisects Angle CBE and CO bisects Angle BCD}$

$\textbf {To Prove => Angle BOC = 90° - 1/2 Angle BAC}$

Solution =>

$= > angle1 + angle2 + y = 180 \: (linear \: pair) \\ put \: angle2 = angle1 \: (given) \\ = > angle1 + angle1 + y = 180 \\ = > 2 \: (angle1) + y = 180 \\ = > 2 \: (angle1) = 180 - y \\ = > angle1 = \frac{180 - y}{2} \\ = > angle1 = \frac{180}{2} - \frac{y}{2} \\ = > angle1 = 90 - \frac{y}{2} \\ \\ = > angle3 + angle4 + z = 180 \\ put \: angle4 = angle3 \\ = > angle3 + angle3 + z = 180 \\ = > 2 \: (angle3) + z = 180 \\ = > 2 \: (angle3) = 180 - z \\ = > angle3 = \frac{180 - z}{2} \\ = > angle3 = \frac{180}{2} - \frac{z}{2} \\ = > angle3 = 90 - \frac{z}{2} \\ \\ = > angle \: boc + angle1 + angle3 = 180 \\ ( angle \: sum \: property) \\ = > angle \: boc + 90 - \frac{y}{2} + 90 - \frac{z}{2} = 180 \\ = > 180 + angle \: boc = 180 + \frac{y}{2} + \frac{z}{2} \\ = > angle \: boc = \frac{1}{2} (y + z) \\ adding \: \frac{1}{2} x \: on \: both \: sides \: as \: we \: need \: this \: is \: our \: answer. \\ = > angle \: boc + \frac{1}{2} x = \frac{1}{2} x + \frac{1}{2} z + \frac{1}{2} y \\ = > angle \: boc + \frac{1}{2} x = \frac{1}{2} (x + y + z) \\ = > angle \: boc + \frac{1}{2} x = \frac{1}{2} \times 180 \\ = > angle \: boc = 90 - \frac{1}{2} x \\ put \: x = angle \: bac \\ = > angle \: boc = 90 - \frac{1}{2} angle \: bac \\ \\ hence \: proved$

$\text{Thanks!!}$