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HiSET: Math : Angle measure, central angles, and inscribed angles

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Example Question #71 : Hi Set: High School Equivalency Test: Math

$\angle 1$ $\displaystyle \angle 1$ and $\angle 2$ $\displaystyle \angle 2$ are a pair of vertical angles.

$\angle 3$ $\displaystyle \angle 3$ and $\angle 4$ $\displaystyle \angle 4$ are a linear pair.

$\angle 2$ $\displaystyle \angle 2$ and $\angle 3$ $\displaystyle \angle 3$ are the two acute angles of a right triangle.

Which of the following must be true?

Possible Answers:

$m \angle 1 = m \angle 4$ $\displaystyle m \angle 1 = m \angle 4$

$m \angle 1 + 90^{\circ }= m \angle 4$ $\displaystyle m \angle 1 + 90^{\circ }= m \angle 4$

$m \angle 1 +m \angle 4 = 90^{\circ }$ $\displaystyle m \angle 1 +m \angle 4 = 90^{\circ }$

$m \angle 1 +m \angle 4 = 180^{\circ }$ $\displaystyle m \angle 1 +m \angle 4 = 180^{\circ }$

$m \angle 4 + 90^{\circ }= m \angle 1$ $\displaystyle m \angle 4 + 90^{\circ }= m \angle 1$

Correct answer:

$m \angle 1 + 90^{\circ }= m \angle 4$ $\displaystyle m \angle 1 + 90^{\circ }= m \angle 4$

Explanation:

$\angle 1$ $\displaystyle \angle 1$ and $\angle 2$ $\displaystyle \angle 2$ are a pair of vertical angles; it follows that

$m \angle 1 = m \angle 2$ $\displaystyle m \angle 1 = m \angle 2$

$\angle 3$ $\displaystyle \angle 3$ and $\angle 4$ $\displaystyle \angle 4$ are a linear pair; it follows that they are supplementary - that is,

$m \angle 3 + m \angle 4 = 180^{\circ }$ $\displaystyle m \angle 3 + m \angle 4 = 180^{\circ }$.

$\angle 2$ $\displaystyle \angle 2$ and $\angle 3$ $\displaystyle \angle 3$ are the two acute angles of a right triangle; it follows that they are complementary - that is,

$m \angle 2 + m \angle 3 = 90^{\circ }$ $\displaystyle m \angle 2 + m \angle 3 = 90^{\circ }$.

Therefore, we have the three statements

$m \angle 1 = m \angle 2$ $\displaystyle m \angle 1 = m \angle 2$

$m \angle 2 + m \angle 3 = 90^{\circ }$ $\displaystyle m \angle 2 + m \angle 3 = 90^{\circ }$

$m \angle 3 + m \angle 4 = 180^{\circ }$ $\displaystyle m \angle 3 + m \angle 4 = 180^{\circ }$

From the second statement, we can subtract $m \angle 2$ $\displaystyle m \angle 2$ from both sides to get

$m \angle 2 + m \angle 3 - m \angle 2 = 90^{\circ } - m \angle 2$ $\displaystyle m \angle 2 + m \angle 3 - m \angle 2 = 90^{\circ } - m \angle 2$

$m \angle 3 = 90^{\circ } - m \angle 2$ $\displaystyle m \angle 3 = 90^{\circ } - m \angle 2$

Substitute this expression for $m \angle 3$ $\displaystyle m \angle 3$ in the third expression to get

$90^{\circ } - m \angle 2 + m \angle 4 = 180^{\circ }$ $\displaystyle 90^{\circ } - m \angle 2 + m \angle 4 = 180^{\circ }$

Substitute $m \angle 1$ $\displaystyle m \angle 1$ for $m \angle 2$ $\displaystyle m \angle 2$:

$90^{\circ } - m \angle 1 + m \angle 4 = 180^{\circ }$ $\displaystyle 90^{\circ } - m \angle 1 + m \angle 4 = 180^{\circ }$

Add $m \angle 1 - 90^{\circ }$ $\displaystyle m \angle 1 - 90^{\circ }$ to both sides:

$90^{\circ } - m \angle 1 + m \angle 4 +m \angle 1 - 90^{\circ }= 180^{\circ }+m \angle 1 - 90^{\circ }$ $\displaystyle 90^{\circ } - m \angle 1 + m \angle 4 +m \angle 1 - 90^{\circ }= 180^{\circ }+m \angle 1 - 90^{\circ }$

$m \angle 4 = 90^{\circ }+m \angle 1$ $\displaystyle m \angle 4 = 90^{\circ }+m \angle 1$,

or, rearranged,

$m \angle 1 + 90^{\circ }= m \angle 4$ $\displaystyle m \angle 1 + 90^{\circ }= m \angle 4$.

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HiSET: Math : Angle measure, central angles, and inscribed angles

Study concepts, example questions & explanations for HiSET: Math

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Example Question #71 : Hi Set: High School Equivalency Test: Math

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