Question

ABC is a triangle inscribed in a circle with centre o. If angle AOC=130 degrees and angleBOC=150 degrees. Find angleACB

Answer 1

The value of $\angle ACB$ is $40\ degree$

Step-by-step explanation:

Given,

$\triangle ABC$ inscribed in a circle,$\angle AOC=130\ degree$ and $\angle BOC=150\ degree$

From Figure,

$OA=OB=OC=r$ where $r=$radius of the circle.

In $\triangle OAC$,

$\angle OAC=\angle OCA$ (∵$OA=OC$)

∴$2\angle OCA+130=180$

⇒$2\angle OCA=180-130$

⇒$\angle OCA=\frac{50}{2}$

∴ $\angle OCA=25\ degree$

Also in $\triangle OBC$

$\angle OBC=\angle OCB$ (∵$OB=OC$)

∴ $2\angle OCB+150=180$

⇒$2\angle OCB=30$

⇒$\angle OCB=\frac{30}{2}$

∴ $\angle OCB=15\ degree$

∴ $\angle ACB=\angle OCA+\angle OCB$

⇒$\angle ACB=25+15$

⇒$\angle ACB=40\ degree$

So, The value of $\angle ACB$ is $40\ degree$