Angles - GED Math

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Question

Give the number of sides of a regular polygon whose interior angles have measure .

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Answer

The easiest way to solve this is to look at the exterior angles, each of which have measure . Since each exterior angle of a regular polygon with sides is $\frac{360}{N}$ , we solve for in the following equation:

$$\frac{360}{N}$ = 10$

$\frac{360}{N}$ \cdot $\frac{N }{10}$ = 10 \cdot $\frac{N }{10}$

The polygon has 36 sides.

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Question

What is the measure of each angle of a regular octagon?

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Answer

The sum of the degree measures of the angles of a polygon with sides is . Since an octagon has eight sides, substitute to get:

$180(8-2 ) = 180 \times6 = 1,080$

Each angle of a regular polygon has equal measure, so divide this by 8 to get the measure of one angle:

$1,080 \div 8 = 135$ , the degree measure of one angle.

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Question

Give the measure of each interior angle of a regular 72-sided polygon.

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Answer

A regular polygon with sides has interior angles of measure each. Substitute 72 for .

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Question

Note: Figure NOT drawn to scale.

The above hexagon is regular. What is ?

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Answer

Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures

.

The four angles of the quadrilateral are $120 ^{\circ }, $120^{\circ }$, $45^{\circ }$, $x^{\circ }$$ . Their sum is , so we can set up an equation and solve for :

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Question

The above octagon is regular. What is ?

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Answer

Three of the angles of the pentagon formed are angles of a regular octagon, so each measures

.

The five angles of the pentagon are . Their sum is , so we can set up an equation and solve for :

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Question

Refer to the above diagram.

Which of these is a valid alternative name for $\angle FEC$ ?

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Answer

When naming an angle after three points, the middle letter must be its vertex, or the point at which its sides meet - this is . The other two letters must refer to points on its two sides. Therefore, $\angle FEC$ includes on one side, making one of its sides $\overrightarrow{EF}$ , and on the other, making the other side $\overrightarrow{EC}$ .

An alternative name for this angle must be one of two things:

It can be named only after its vertex - that is, $\angle E$ - but only if there is no ambiguity as to which angle is being named. Since more than one angle in the diagram has vertex , $\angle E$ is not a correct choice.

It can be named after three points. Again, the middle letter must be vertex , so we can throw out $\angle GCE$ and $\angle GFE$ .

The only possible choice is $\angle CEA$ .

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Question

Refer to the above figure. $\Delta OKR$ is equilateral, and Quadrilateral is a square.

Evaluate $m \angle ROP$ .

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Answer

By angle addition,

$m \angle ROP = m \angle ROK + m \angle KOP$ .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$\angle KOP$ , as an angle of a square, has measure $90 ^{\circ }$ .

Therefore,

$m \angle ROP = 60 ^{\circ } + $90^{\circ }$ = 150 ^{\circ }$ .

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Question

Refer to the above figure, which shows Square and regular Pentagon .

Evaluate $m \angle NOQ$ .

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Answer

By angle addition,

$m \angle NOQ = m \angle NOK + m \angle KOQ$ .

$\angle NOK$ is an angle of a regular pentagon and has measure .

$\angle KOQ$ is one of two acute angles of isosceles right triangle $\Delta KOQ$ , so $m \angle KOQ = $45^{\circ }$$ .

$m \angle NOQ = $108^{\circ }$ + $45^{\circ }$ = $153^{\circ }$$

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. $\Delta OKR$ is equilateral and Pentagon is regular.

Evaluate $m \angle NRO$ .

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Answer

First, we find $m \angle NOR$ .

By angle addition,

$m \angle NOR + m \angle ROK = m \angle NOK$ .

$\angle NOK$ is an angle of a regular pentagon and has measure .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$m \angle NOR + 60 ^{\circ }= 108 ^{\circ }$

$m \angle NOR = 48 ^{\circ }$

$\Delta OKR$ is equilateral, so $\overline{OR}$ \cong $\overline{OK}$ ; Pentagon is regular, so $\overline{OK}$ \cong $\overline{NO}$ . Therefore, $\overline{OR}$ \cong $\overline{ON}$ , and by the Isosceles Triangle Theorem, $m \angle NRO = m \angle RNO$ .

The degree measures of three angles of a triangle total $180 ^{\circ }$ , so:

$m \angle NRO + m \angle RNO + m \angle NOR = $180^{\circ }$$

$m \angle NRO + m \angle NRO + $48^{\circ }$ = $180^{\circ }$$

$2 m \angle NRO + $48^{\circ }$ = $180^{\circ }$$

$2 m \angle NRO = $132^{\circ }$$

$m \angle NRO = $66^{\circ }$$

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Question

Three consecutive even angles add up to . What must be the value of the second largest angle?

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Answer

Let be an even angle. The next consecutive even values are .

Set up an equation such that all angles added equal to 180.

Divide by three on both sides.

$\frac{3x}{3}$=\frac{174}{3}$

The second largest angle is .

Substitute the value of in to the expression.

The answer is:

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Question

What is the measure of each angle of a regular octagon?

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Answer

The sum of the degree measures of the angles of a polygon with sides is . Since an octagon has eight sides, substitute to get:

$180(8-2 ) = 180 \times6 = 1,080$

Each angle of a regular polygon has equal measure, so divide this by 8 to get the measure of one angle:

$1,080 \div 8 = 135$ , the degree measure of one angle.

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Question

Give the number of sides of a regular polygon whose interior angles have measure .

Tap to reveal answer

Answer

The easiest way to solve this is to look at the exterior angles, each of which have measure . Since each exterior angle of a regular polygon with sides is $\frac{360}{N}$ , we solve for in the following equation:

$$\frac{360}{N}$ = 10$

$\frac{360}{N}$ \cdot $\frac{N }{10}$ = 10 \cdot $\frac{N }{10}$

The polygon has 36 sides.

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Question

Give the measure of each interior angle of a regular 72-sided polygon.

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Answer

A regular polygon with sides has interior angles of measure each. Substitute 72 for .

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Question

Note: Figure NOT drawn to scale.

The above hexagon is regular. What is ?

Tap to reveal answer

Answer

Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures

.

The four angles of the quadrilateral are $120 ^{\circ }, $120^{\circ }$, $45^{\circ }$, $x^{\circ }$$ . Their sum is , so we can set up an equation and solve for :

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Question

The above octagon is regular. What is ?

Tap to reveal answer

Answer

Three of the angles of the pentagon formed are angles of a regular octagon, so each measures

.

The five angles of the pentagon are . Their sum is , so we can set up an equation and solve for :

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Question

Refer to the above diagram.

Which of these is a valid alternative name for $\angle FEC$ ?

Tap to reveal answer

Answer

When naming an angle after three points, the middle letter must be its vertex, or the point at which its sides meet - this is . The other two letters must refer to points on its two sides. Therefore, $\angle FEC$ includes on one side, making one of its sides $\overrightarrow{EF}$ , and on the other, making the other side $\overrightarrow{EC}$ .

An alternative name for this angle must be one of two things:

It can be named only after its vertex - that is, $\angle E$ - but only if there is no ambiguity as to which angle is being named. Since more than one angle in the diagram has vertex , $\angle E$ is not a correct choice.

It can be named after three points. Again, the middle letter must be vertex , so we can throw out $\angle GCE$ and $\angle GFE$ .

The only possible choice is $\angle CEA$ .

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Question

Refer to the above figure. $\Delta OKR$ is equilateral, and Quadrilateral is a square.

Evaluate $m \angle ROP$ .

Tap to reveal answer

Answer

By angle addition,

$m \angle ROP = m \angle ROK + m \angle KOP$ .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$\angle KOP$ , as an angle of a square, has measure $90 ^{\circ }$ .

Therefore,

$m \angle ROP = 60 ^{\circ } + $90^{\circ }$ = 150 ^{\circ }$ .

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Question

Refer to the above figure, which shows Square and regular Pentagon .

Evaluate $m \angle NOQ$ .

Tap to reveal answer

Answer

By angle addition,

$m \angle NOQ = m \angle NOK + m \angle KOQ$ .

$\angle NOK$ is an angle of a regular pentagon and has measure .

$\angle KOQ$ is one of two acute angles of isosceles right triangle $\Delta KOQ$ , so $m \angle KOQ = $45^{\circ }$$ .

$m \angle NOQ = $108^{\circ }$ + $45^{\circ }$ = $153^{\circ }$$

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. $\Delta OKR$ is equilateral and Pentagon is regular.

Evaluate $m \angle NRO$ .

Tap to reveal answer

Answer

First, we find $m \angle NOR$ .

By angle addition,

$m \angle NOR + m \angle ROK = m \angle NOK$ .

$\angle NOK$ is an angle of a regular pentagon and has measure .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$m \angle NOR + 60 ^{\circ }= 108 ^{\circ }$

$m \angle NOR = 48 ^{\circ }$

$\Delta OKR$ is equilateral, so $\overline{OR}$ \cong $\overline{OK}$ ; Pentagon is regular, so $\overline{OK}$ \cong $\overline{NO}$ . Therefore, $\overline{OR}$ \cong $\overline{ON}$ , and by the Isosceles Triangle Theorem, $m \angle NRO = m \angle RNO$ .

The degree measures of three angles of a triangle total $180 ^{\circ }$ , so:

$m \angle NRO + m \angle RNO + m \angle NOR = $180^{\circ }$$

$m \angle NRO + m \angle NRO + $48^{\circ }$ = $180^{\circ }$$

$2 m \angle NRO + $48^{\circ }$ = $180^{\circ }$$

$2 m \angle NRO = $132^{\circ }$$

$m \angle NRO = $66^{\circ }$$

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Question

Three consecutive even angles add up to . What must be the value of the second largest angle?

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Answer

Let be an even angle. The next consecutive even values are .

Set up an equation such that all angles added equal to 180.

Divide by three on both sides.

$\frac{3x}{3}$=\frac{174}{3}$

The second largest angle is .

Substitute the value of in to the expression.

The answer is:

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