Question

In the figure measure ∠AOC

• A. 150^o
• B. 160^o
• C. 140^o
• D. 180^o

Answer 1

The correct option is C

140^o

$\angle$ABC = $X^{\circ }$

$\angle$AOC = $2X^{\circ }$

$\angle$OAC + $\angle$OCA = ${180}^{\circ }$ - $2X^{\circ }$

$\angle$BAC + $\angle$ACB = ${180}^{\circ }$ - $2X^{\circ }$ + ${30}^{\circ }$ + ${40}^{\circ }$

Sum of angles of a triangle = ${180}^{\circ }$

Therefore,

$\angle$BAC + $\angle$ACB + $\angle$ABC = ${180}^{\circ }$

${180}^{\circ }$ - $2X^{\circ }$ + ${30}^{\circ }$ + ${40}^{\circ }$ + $X^{\circ }$ = ${180}^{\circ }$

X = ${70}^{\circ }$

$\angle$AOC = 2X = ${140}^{\circ }$