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SAT Math : Sectors

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Example Question #21 : Sectors

Secant

Refer to the above diagram. Evaluate the measure of $\angle BND$ .

Possible Answers:

$60^{\circ }$

$45 ^{\circ }$

$30^{\circ }$

$52 \frac{1}{2}^{\circ }$

$37 \frac{1}{2}^{\circ }$

Correct answer:

$45 ^{\circ }$

Explanation:

The total measure of the arcs that comprise a circle is $360 ^{\circ }$ , so from the above diagram,

$m \overarc { AC}+m \overarc { CD}+ m \overarc { DB} + m \overarc {BA } = 360 ^{\circ }$

Substituting the appropriate expression for each arc measure:

t+t+3t+3t = 360

8t = 360

$8t \div 8 = 360 \div 8$

t = 45

Therefore,

$m \overarc { AC}= 45^{\circ }$

and

$m \overarc { DB} = 3 t^{\circ } = 3 \cdot 45^{\circ } = 135^{\circ }$

The measure of the angle formed by the tangent segments $\overline{NB}$ and $\overline{ND}$ , which is $\angle BND$ , is half the difference of the measures of the arcs they intercept, so

$m\angle BND = \frac{1}{2} (m \overarc {DB} - m \overarc {AC})$

Substituting:

$m\angle BND = \frac{1}{2} (135 ^{\circ }- 45^{\circ } )$

$m\angle BND = \frac{1}{2} (90^{\circ } )$

$m\angle BND = 45 ^{\circ }$

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Example Question #11 : How To Find The Angle Of A Sector

Inscribed quad

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Possible Answers:

u= 102

The question cannot be answered from the information given.

u= 92

u= 88

u= 96

Correct answer:

u= 92

Explanation:

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle B$ and $\angle D$ are two such angles, so

$m \angle B + m \angle D = 180 ^{\circ }$

Setting $m \angle B = 88^{\circ }$ and $m \angle D =u ^{\circ }$ , and solving for :

88+ u = 180

88+ u - 88 = 180 - 88

u= 92 ,

the correct response.

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Example Question #12 : How To Find The Angle Of A Sector

Inscribed quad

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Possible Answers:

The question cannot be answered from the information given.

Correct answer:

The question cannot be answered from the information given.

Explanation:

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (180-t)^{\circ }$ , and solving for :

t + (180-t)= 180

180+t-t= 180

180= 180 ,

The statement turns out to be true regardless of the value of . Therefore, without further information, the value of cannot be determined.

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Example Question #13 : How To Find The Angle Of A Sector

Inscribed quad

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Possible Answers:

t= 84

t = 86

t = 80

t = 82

t= 78

Correct answer:

t= 84

Explanation:

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (t+12)^{\circ }$ , and solving for :

t + (t+12 )= 180

2t +12 = 180

2t +12 - 12 = 180 - 12

2t = 168

$2t \div 2 = 168 \div 2$

t = 84 ,

the correct response.

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Example Question #14 : How To Find The Angle Of A Sector

Secant 2

Figure NOT drawn to scale.

Refer to the above diagram. $\overline{AB}$ is a diameter. Evaluate $m \angle CBA$

Possible Answers:

$28^{\circ }$

$56^{\circ }$

$45^{\circ }$

$31^{\circ }$

$62^{\circ }$

Correct answer:

$31^{\circ }$

Explanation:

$\overline{AB}$ is a diameter, so $\overarc{ACB}$ is a semicircle - therefore, $m \overarc{ACB} = 180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc{AC}+ m \overarc{CB} = 180 ^{\circ }$

If we let $t ^{\circ }= m \overarc{AC}$ , then

$t ^{\circ }+ m \overarc{CB} = 180 ^{\circ }$ ,

and

$m \overarc{CB} = (180 -t )^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NC}$ and $\overline{NB}$ are such segments intercepting $\overarc{AC}$ and $\overarc{CB}$ , it holds that

$\frac{1}{2}\left ( m \overarc{CB}- m \overarc{AC} \right ) = m \angle CNB$

Setting $m \overarc{AC} = t ^{\circ }$ , $m \overarc{CB} = (180 -t )^{\circ }$ , and $m \angle CNB = 28 ^{\circ }$ :

$\frac{1}{2} [ \left ( 180-t \right ) - t ] = 28$

$\frac{1}{2} \left ( 180-2 t \right ) = 28$

90-t = 28

90-t + t - 28 = 28 + t - 28

62 = t

$m \overarc{AC} =62^{\circ }$

The inscribed angle that intercepts this arc, $\angle CBA$ , has half this measure:

$m \angle CBA = \frac{1}{2} \cdot m \overarc{AC} = \frac{1}{2} \cdot 62^{\circ } = 31^{\circ }$ .

This is the correct response.

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Example Question #15 : How To Find The Angle Of A Sector

Secant 3 Figure NOT drawn to scale.

In the above figure, $\overline{AB}$ is a diameter. Also, the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5. Give the measure of $\angle TNA$ .

Possible Answers:

$18^{\circ }$

$15^{\circ }$

$21 ^{\circ }$

$24 ^{\circ }$

The measure of $\angle TNA$ cannot be determine from the information given.

Correct answer:

$15^{\circ }$

Explanation:

$\overline{AB}$ is a diameter, so $\overarc {ATB}$ is a semicircle, which has measure $180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc {AT } + m \overarc {BT} = m \overarc {ATB}$

If we let $t ^{\circ }= m \overarc {AT }$ , then, substituting:

$t ^{\circ }+ m \overarc {BT} = 180 ^{\circ }$ ,

and

$m \overarc {BT} =( 180 - t ) ^{\circ }$

the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5; this is also the ratio of their degree measures; that is,

$\frac{m \overarc{BT}}{m \overarc{AT}} = \frac{5}{3}$

Setting $m \overarc {AT } = t ^{\circ }$ and $m \overarc {BT} =( 180 - t ) ^{\circ }$ :

$\frac{180-t }{t} = \frac{7}{5}$

Cross-multiply, then solve for :

7t = 5 (180-t)

7t = 900 - 5t

7t + 5t = 900 - 5t + 5t

12t = 900

$\frac{12t }{12}= \frac{900}{12}$

t = 75

$m \overarc {AT } =75^{\circ }$ , and $m \overarc {BT} =( 180 -75 ) ^{\circ } = 105 ^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NB}$ and $\overline{NT}$ are such segments whose angle $\angle TNA$ intercepts $\overarc{AT}$ and $\overarc{BT}$ , it holds that:

$m \angle TNA = \frac{1}{2}\left ( m \overarc{BT}- m \overarc{AT} \right )$

$m \angle TNA = \frac{1}{2}\left ( 105 ^{\circ } - 75^{\circ } \right )$

$m \angle TNA = \frac{1}{2}\left ( 30 ^{\circ } \right )$

$m \angle TNA = 15^{\circ }$

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SAT Math : Sectors

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #21 : Sectors

Example Question #11 : How To Find The Angle Of A Sector

Example Question #12 : How To Find The Angle Of A Sector

Example Question #13 : How To Find The Angle Of A Sector

Example Question #14 : How To Find The Angle Of A Sector

Example Question #15 : How To Find The Angle Of A Sector

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