Question

A, B, C are any points on the circle with centre O. m(arc BC) = 120${}^{\circ }$ and m(arc AB) = 130${}^{\circ }.$
Then, find m(arc ABC), m (arc ACB), and m (arc BAC).

• A. None of the above
• B. 240${}^{\circ }$, 250${}^{\circ }$, 220${}^{\circ }$
• C. 250${}^{\circ }$, 230${}^{\circ }$, 240${}^{\circ }$
• D. 240${}^{\circ }$, 250${}^{\circ }$, 240${}^{\circ }$

Answer 1

The correct option is C 250${}^{\circ }$, 230${}^{\circ }$, 240${}^{\circ }$

By property of sum of measure of arcs, we must have
m(arc ABC) = m(arc AB) + m(arc BC)
$⟹$ m(arc ABC) = 130${}^{\circ }$ + 120${}^{\circ }$
= 250${}^{\circ }$
m(arc AC) = 360${}^{\circ }$ - m(arc ABC)
$⟹$ m(arc AC) = 360${}^{\circ }$ - 250${}^{\circ }$
= 110${}^{\circ }$

m(arc ACB) = m(arc AC) + m(arc BC)
$⟹$ m(arc ACB) = 110${}^{\circ }$ + 120${}^{\circ }$
= 230${}^{\circ }$
m(arc BAC) = 360${}^{\circ }$ - m(arc BC)
= 360${}^{\circ }$ - 120${}^{\circ }$
= 240${}^{\circ }$