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All HiSET: Math Resources

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Flashcards: Angle measure, central angles, and inscribed angles

Pool shape

A quadrilateral is shown, and the angle measures of 3 interior angles are given. Find x, the missing angle measure.

$33^{\circ}$

$228^{\circ}$

$52^{\circ}$

$128^{\circ}$

$94^{\circ}$

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Flashcards: Angle measure, central angles, and inscribed angles

A quadrilateral is shown, and the angle measures of 3 interior angles are given. Find x, the missing angle measure.

Your answer:

$33^{\circ}$

Your answer:

$228^{\circ}$

Your answer:

$52^{\circ}$

Your answer:

$94^{\circ}$

Correct answer:

$128^{\circ}$

Explanation:

The sum of the measures of the interior angles of a quadrilateral is 360 degrees. The sum of the measures of the interior angles of any polygon can be determined using the following formula:

$Sum \ of\ Interior \ Angles = 180^{\circ}(n-2)$ , where is the number of sides.

For example, with a quadrilateral, which has 4 sides, you obtain the following calculation:

$Sum \ of\ Interior \ Angles = 180^{\circ}(4-2)=180^{\circ}\cdot2=360^{\circ}$

Solving for requires setting up an algebraic equation, adding all 4 angles to equal 360 degrees:

$50^{\circ}+67^{\circ}+115^{\circ}+x^{\circ}=360^{\circ}$

Solving for is straightforward: subtract the values of the 3 known angles from both sides:

$x^{\circ}= 360^{\circ}-50^{\circ}-67^{\circ}-115^{\circ}$

$x^{\circ}=128^{\circ}$

Flashcards: Angle measure, central angles, and inscribed angles

Intercepted 2

is the center of the above circle. Calculate $m \angle AOB$ .

$135^{\circ }$

$130^{\circ }$

$125^{\circ }$

$120^{\circ }$

$140^{\circ }$

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Flashcards: Angle measure, central angles, and inscribed angles

is the center of the above circle. Calculate $m \angle AOB$ .

Your answer:

$130^{\circ }$

Your answer:

$125^{\circ }$

Your answer:

$120^{\circ }$

Your answer:

$140^{\circ }$

Correct answer:

$135^{\circ }$

Explanation:

$\angle AOB$ is the central angle that intercepts $\overarc{AB}$ , so

$m \angle AOB = m \overarc{AB}$ .

Therefore, we need to find $m \overarc{AB}$ to obtain our answer.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

360- 2 x = 90

Subtract 360 from both sides:

360- 2 x - 360 = 90 - 360

-2x = -270

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

x = 135 ,

the measure of $\overarc{AB}$ and, consequently, that of $\angle AOB$ .

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Flashcards: Angle measure, central angles, and inscribed angles

Flashcards: Angle measure, central angles, and inscribed angles

Flashcards: Angle measure, central angles, and inscribed angles

Flashcards: Angle measure, central angles, and inscribed angles

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Flashcards: Angle measure, central angles, and inscribed angles

Flashcards: Angle measure, central angles, and inscribed angles

Flashcards: Angle measure, central angles, and inscribed angles

Flashcards: Angle measure, central angles, and inscribed angles

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