Answer:
Solution given:
14.
34x+55x+2=180°[sum of opposite angle of a cyclic quadrilateral is supplementary]
89x=180°-2
x=178/89
x=2°
<u>now</u>
<u>m</u><u> </u><u>arc</u><u> </u><u>VX</u><u>=</u><u>2</u><u>×</u><u> </u><u><</u><u>H</u><u>[</u><u> </u><u>central</u><u> </u><u>angle</u><u> </u><u>is</u><u> </u><u>twice</u><u> </u><u>the</u><u> </u><u>inscribed</u><u> </u><u>angle</u><u> </u><u>standing</u><u> </u><u>on</u><u> </u><u>a</u><u> </u><u>same</u><u> </u><u>base</u><u>]</u>
m arc VX=2(34×2)=<u>1</u><u>3</u><u>6</u><u>°</u>
<u>Option</u><u> </u><u>C</u><u>.</u>
15.
Solution given:
<B=80°
<C=½ arc AB[central angle is twice the inscribed angle standing on a same base]
<C=½×120°=60°
we have
<A+<B+<C=180°[ sum of interior angle of a triangle is 180 ]
<A=180-80°-60°
<A=<u>4</u><u>0</u><u>°</u>
<u>o</u><u>p</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>C</u><u>.</u>