HiSET - 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.

$128^{\circ}$

$33^{\circ}$

$52^{\circ}$

$228^{\circ}$

$94^{\circ}$

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:

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

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

$\angle 1$ and $\angle 2$ are complementary angles.

$\angle 1$ and $\angle 3$ are supplementary angles.

$\angle 2 \cong \angle 4$

$m \angle 3= 112^{\circ }$

Evaluate $m \angle 4$ .

$22^{\circ }$

$68^{\circ }$

$112^{\circ }$

$158^{\circ }$

$90^{\circ }$

Explanation

$\angle 1$ and $\angle 3$ are supplementary angles, so, by definition,

$m \angle 1 + m \angle 3 = 180^{\circ }$

$m \angle 3= 112^{\circ }$ , so substitute and solve for $m \angle 1$ :

$m \angle 1 + 112^{\circ } = 180^{\circ }$

$m \angle 1 + 112^{\circ } -112^{\circ } = 180^{\circ }-112^{\circ }$

$m \angle 1 = 68 ^{\circ }$

$\angle 1$ and $\angle 2$ are complementary angles, so, by definition,

$m \angle 1 + m \angle 2 = 90^{\circ }$

Substitute and solve for $m \angle 2$ :

$68 ^{\circ } + m \angle 2 = 90^{\circ }$

$68 ^{\circ } + m \angle 2 - 68 ^{\circ } = 90^{\circ }-68 ^{\circ }$

$m \angle 2 =22^{\circ }$

$\angle 2 \cong \angle 4$ - that is, the angles have the same measure. Therefore,

$m \angle 4 =22^{\circ }$ .

Intercepted 2

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

$135^{\circ }$

$120^{\circ }$

$130^{\circ }$

$125^{\circ }$

$140^{\circ }$

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:

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

Subtract 360 from both sides:

Divide both sides by :

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

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

Inscribed nonagon

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

$72^{\circ }$

$75^{\circ }$

$80^{\circ }$

$76^{\circ }$

$84^{\circ }$

Explanation

Examine the diagram below, which divides $\angle AOB$ into two congruent angles, one of which is $\angle 1$ :

Inscribed nonagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle AOB$ has measure twice this; that is,

$m \angle AOB = 2 \times 36 ^{\circ } = 72 ^{\circ }$ .

Inscribed nonagon

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle OBC$ .

$36 ^{\circ }$

$40^{\circ }$

$32^{\circ }$

$42^{\circ }$

$45^{\circ }$

Explanation

Consider the triangle $\bigtriangleup OBC$ . Since $\overline{OB}$ and $\overline{OC}$ are radii, they are congruent, and by the Isosceles Triangle Theorem, $m \angle OBC = m \angle OCB$ .

Now, examine the figure below, which divides $\angle BOC$ into three congruent angles, one of which is $\angle 1$ :

Inscribed nonagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle BOC$ has measure three times this; that is,

$m \angle BOC = 3 \times 36 ^{\circ } = 108 ^{\circ }$ .

The measures of the interior angles of a triangle total $108^{\circ }$ , so

$m \angle OCB + m \angle OBC+m \angle BOC = 180$

Substituting 108 for $\angle BOC$ and $m \angle OBC$ for $m \angle OCB$ :

$m \angle OBC+ m \angle OBC+108 = 180$

$2 m \angle OBC+108 = 180$

$2 m \angle OBC+108- 108 = 180- 108$

$2 m \angle OBC=72$

$2 m \angle OBC \div 2 =72 \div 2$

$m \angle OBC= 36$

Inscribed heptagon

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

$154 \frac{2}{7}^{\circ }$

$155 \frac{1}{7}^{\circ }$

$156 \frac{4}{7}^{\circ }$

$152 \frac{6}{7}^{\circ }$

$153 \frac{5}{7}^{\circ }$

Explanation

Examine the diagram below, which divides $\angle AOB$ into three congruent angles, one of which is $\angle 1$ :

Inscribed heptagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure three times this; that is,

$m \angle AOC = 3 \times51 \frac{3}{7}^{\circ } = 154 \frac{2}{7}^{\circ }$

Inscribed nonagon

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle CBE$ .

$36^{\circ }$

$32^{\circ }$

$30^{\circ }$

$40^{\circ }$

$42^{\circ }$

Explanation

Through symmetry, it can be seen that Quadrilateral is a trapezoid, such that $\overline{BC} || \overline{DE }$ . By the Same-Side Interior Angle Theorem, $\angle CBE$ and $\angle DEB$ are supplementary - that is,

$m \angle CBE +\angle DEB = 180$ .

The measure of $\angle DEB$ can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where :

$m \angle DEB= \frac{180^{\circ } (10-2)}{10} = \frac{180^{\circ } (8)}{10} = \frac{1,440}{10}^{\circ } =144 ^{\circ }$

Substituting:

$m \angle CBE +144= 180$

$m \angle CBE +144 - 144 = 180 - 144$

$m \angle CBE = 36$

Inscribed heptagon

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOC$ .

$102 \frac{6}{7}^{\circ }$

$104 \frac{2}{7}^{\circ }$

$103 \frac{5}{7}^{\circ }$

$105 \frac{1}{7}^{\circ }$

$101 \frac{3}{7}^{\circ }$

Explanation

Examine the diagram below, which divides $\angle AOC$ into two congruent angles, one of which is $\angle 1$ :

Inscribed heptagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting , the measure of $\angle 1$ is