In figure, AOB is a straight …
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Posted by Brizund Momin 5 years, 5 months ago
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Sia ? 5 years, 5 months ago
AOB is a straight line
{tex}\therefore{/tex} {tex}\angle{/tex}AOC + {tex}\angle{/tex}COD + {tex}\angle{/tex}DOB = 180°
{tex}\Rightarrow{/tex} (3x – 5)° + 55° + (x + 20)° = 180°
{tex}\Rightarrow{/tex} 3x – 5° + 55° + x + 20° = 180°
{tex}\Rightarrow{/tex} 4x + 70° = 180°
{tex}\Rightarrow{/tex} 4x = 180° – 70°
{tex}\Rightarrow{/tex} 4x = 110°
{tex}\Rightarrow{/tex} x =110/4 = 27 ½°
Therefore, {tex}\angle{/tex}AOC = 3x – 5° = 3 {tex}\times{/tex} 27½° – 5 = 77½°
and {tex}\angle{/tex}BOD = x + 20 = 27 ½ + 20 = 47 ½o
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