Question

Question
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Answer 1

$\Large\textrm{Note}$

Read the explanation along with the drawing, step-by-step.

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$\Large\textrm{Explanation}$

In the figure, we observe $\overline{\rm{AB}}\parallel\overline{\rm{CD}}$.

$\rm\therefore\angle ABC=\underline{\angle BCD=34^{\circ}}$

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Join the points to form a new triangle $\rm\triangle ACE$, $\rm\triangle ABE$ and $\rm\triangle BDE$.

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In $\rm\triangle ABE$, the central angle is linear. Hence, the circumference angle is half, which is $\rm\angle AEB=90^{\circ}$.

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If an arc is common, the central angle is the same, hence the circumference angle.

$\rm\therefore\angle ABC=\underline{\angle AEC=34^{\circ}}$

$\rm\therefore\angle BCD=\underline{\angle BED=34^{\circ}}$

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$\rm x+\angle AEC+\angle BED=90^{\circ}$

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$\rm x+34^{\circ}+34^{\circ}=90^{\circ}$

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$\rm x+68^{\circ}=90^{\circ}$

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$\rm\therefore x=22^{\circ}$

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$\Large\textrm{Learn More}$

$\textbf{- Central Angle Theorem}$

The central angle is always double the circumference angle.

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As an arc shares the central angle, it becomes the reason the circumference angle on an arc is equal.