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ISEE Upper Level Quantitative : How to find the angle of a sector

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

Inscribed angle 2

In the above figure, $\overline{AD}$ is a diameter of the circle.

Which is the greater quantity?

(a) $m \angle CAD$

(b) $m \angle CBD$

Possible Answers:

(a) and (b) are equal

(b) is the greater quantity

It is impossible to determine which is greater from the information given

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

That $\overline{AD}$ is a diameter of the circle is actually irrelevant to the problem. Two inscribed angles of a circle that both intercept the same arc, as $\angle CAD$ and $\angle CBD$ both do here, have the same measure.

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

$\bigtriangleup ABC$ is inscribed in a circle. $\overarc {ABC }$ is a semicircle. $m \angle A = 40 ^{\circ }$ .

Which is the greater quantity?

(a) $m \overarc {BAC}$

(b) $m \overarc {BCA }$

Possible Answers:

(b) is the greater quantity

It is impossible to determine which is greater from the information given

(a) is the greater quantity

(a) and (b) are equal

Correct answer:

(a) is the greater quantity

Explanation:

The figure referenced is below:

Inscribed angle

$\overarc {ABC }$ is a semicircle, so $\overarc {ADC }$ is one as well; as a semicircle, its measure is $180 ^{\circ }$ . The inscribed angle that intercepts this semicircle, $\angle B$ , is a right angle, of measure $90 ^{\circ }$ . $m \angle A = 40 ^{\circ }$ , and the sum of the measures of the interior angles of a triangle is $180 ^{\circ }$ , so

$m \angle C = 180 ^{\circ } - (m \angle A + m \angle B )$

$= 180 ^{\circ } - (40 ^{\circ } +90 ^{\circ } )$

$= 180 ^{\circ } - 130 ^{\circ }$

$= 50 ^{\circ }$

$\angle C$ has greater measure than $\angle A$ , so the minor arc intercepted by $\angle C$ , which is $\overarc {BA}$ , has greater measure than that intercepted by $\angle A$ , which is $\overarc {BC}$ . It follows that the major arc corresponding to the latter, which is $\overarc {BA C}$ , has greater measure than that corresponding to the former, which is $\overarc {BC A }$ .

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Example Question #191 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Inscribed angle 3

In the above figure, is the center of the circle, and $m \overarc {AD} = 60 ^{\circ }$ . Which is the greater quantity?

(a)

(b)

Possible Answers:

(a) is the greater quantity

(b) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

Correct answer:

(a) is the greater quantity

Explanation:

Construct $\overline{OD}$ . The new figure is below:

Inscribed angle 3

$m \overarc {AD} = 60 ^{\circ }$ , so $m \overarc {AC} < 60 ^{\circ }$ . It follows that their respective central angles have measures

$m \angle AOD = 60 ^{\circ }$

and

$m \angle AOC < 60 ^{\circ }$ .

Also, since $m \overarc {AD} = 60 ^{\circ }$ and $m \overarc {ACB} = 180 ^{\circ }$ - $\overarc {ACB}$ being a semicircle - by the Arc Addition Principle, $m \overarc {BD} = 120^{\circ }$ . $\angle DAO$ , an inscribed angle which intercepts this arc, has half this measure, which is $60 ^{\circ }$ . The other angle of $\bigtriangleup ADO$ , which is $\angle ADO$ , also measures $60 ^{\circ }$ , so $\bigtriangleup ADO$ is equilateral.

OD = OC , since all radii are congruent;

OA = OA by reflexivity;

$m \angle AOC < m \angle AOD$

By the Side-Angle-Side Inequality Theorem (or Hinge Theorem), it follows that AC < AD . Since $\bigtriangleup ADO$ is equilateral, AD = OD , and since all radii are congruent, OD = OC . Substituting, it follows that AC < OC .

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Example Question #21 : Sectors

Trapezoid ABCD is inscribed in a circle, with $\overline{AD}$ a diameter.

Which is the greater quantity?

(a) $m \overarc{AC}$

(b) $m \overarc{DB}$

Possible Answers:

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

Below is the inscribed trapezoid referenced, along with its diagonals.

Inscribed angle 3

$\overline{AD} ||\overline{BC}$ , so, by the Alternate Interior Angles Theorem,

$\angle ACB \cong \angle CAD$ , and their intercepted angles are also congruent - that is,

$m \overarc{AB} = m \overarc{DC}$

$m \overarc{AB}+ m \overarc{BC} = m \overarc{DC} + m \overarc{CB}$

By the Arc Addition Principle,

$m \overarc{A C} = m \overarc{D B}$ .

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Example Question #191 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Circle 2

In the above figure, $\overline{AD}$ is a diameter of the circle. Which is the greater quantity?

(a) $m \angle ADB$

(b) $m \angle ACB$

Possible Answers:

(a) and (b) are equal

(a) is the greater quantity

It is impossible to determine which is greater from the information given

(b) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

Both $\angle ADB$ and $\angle ACB$ are inscribed angles of the same circle which intercept the same arc; they are therefore of the same measure. The fact that $\overline{AD}$ is a diameter of the circle is actually irrelevant to the problem.

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Example Question #191 : Plane Geometry

Tangents 1

Figure NOT drawn to scale.

Refer to the above diagram. $m \overarc {NO}$ is the arithmetic mean of $m \overarc {NP}$ and $m \overarc {OP}$ .

Which is the greater quantity?

(a) $m \angle NTO$

(b) $60 ^{\circ }$

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

$m \overarc {NO}$ is the arithmetic mean of $m \overarc {NP}$ and $m \overarc {OP}$ , so

$m \overarc {NO} = \frac{1}{2} \left ( m \overarc {NP} + m \overarc {OP} \right )$

By arc addition, this becomes

$m \overarc {NO} = \frac{1}{2} \cdot m \overarc {NPO}$

Also, $m \overarc {NO} + m \overarc {NPO}= 360 ^{\circ }$ , or, equivalently,

$m \overarc {NPO} = 360 ^{\circ } - m \overarc {NO}$ , so

$m \overarc {NO} = \frac{1}{2} \cdot \left ( 360 ^{\circ } - m \overarc {NO} \right )$

Solving for $m \overarc {NO}$ :

$m \overarc {NO} = \frac{1}{2} \cdot 360 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO}$

$m \overarc {NO} = 180 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO}$

$m \overarc {NO} + \frac{1}{2} \cdot m \overarc {NO} = 180 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO} + \frac{1}{2} \cdot m \overarc {NO}$

$\frac{3}{2} \cdot m \overarc {NO} = 180 ^{\circ }$

$\frac{2} {3} \cdot \frac{3}{2} \cdot m \overarc {NO} = \frac{2} {3} \cdot 180 ^{\circ }$

$m \overarc {NO} = 120 ^{\circ }$

Also,

$m \overarc {NPO} = 360 ^{\circ } - m \overarc {NO} = 360 ^{\circ } -120 ^{\circ } = 240 ^{\circ }$

If two tangents are drawn to a circle, the measure of the angle they form is half the difference of the measures of the arcs they intercept, so

$m \angle NTO = \frac{1}{2}(m \overarc{NPO} - m \overarc{N O} ) = \frac{1}{2}(240^{\circ }- 120^{\circ } ) = \frac{1}{2}(120^{\circ } ) = 60^{\circ }$

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Example Question #17 : How To Find The Angle Of A Sector

Inscribed angle 4

Figure NOT drawn to scale

In the above diagram, $BA = AX \sqrt{2}$ .

Which is the greater quantity?

(a) $m \overarc{AD}$

(b) $m \overarc{BC}$

Possible Answers:

(a) and (b) are equal

(b) is the greater quantity

(a) is the greater quantity

It is impossible to determine which is greater from the information given

Correct answer:

(a) and (b) are equal

Explanation:

$\bigtriangleup AXB$ is a right triangle whose hypotenuse $\overline{AB}$ has length $\sqrt{2}$ times that of leg $\overline{AX}$ . This is characteristic of a triangle whose acute angles both have measure $45 ^{\circ }$ -and consequently, whose acute angles are congruent. Therefore,

$\angle A \cong \angle B$

These inscribed angles being congruent, the arcs they intercept, $\overarc{AD}$ and $\overarc{BC}$ , are also congruent.

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

Inscribed angle 4

Figure NOT drawn to scale

In the above diagram, $m \overarc {BC } = 100^{\circ }$ .

Which is the greater quantity?

(a) $m \angle A$

(b) $m \angle B$

Possible Answers:

(b) is the greater quantity

It is impossible to determine which is greater from the information given

(a) is the greater quantity

(a) and (b) are equal

Correct answer:

(a) is the greater quantity

Explanation:

$\angle A$ is an inscribed angle, so its degree measure is half that of the arc it intercepts, $\overarc {BC }$ :

$m \angle A = \frac{1}{2} \cdot m \overarc {BC } = \frac{1}{2} \cdot 100^{\circ } = 50^{\circ }$ .

$\angle A$ and $\angle B$ are acute angles of right triangle $\bigtriangleup AXB$ . They are therefore complimentary - that is, their degree measures total $90^{\circ }$ . Consequently,

$m \angle B+ m \angle A = 90^{\circ }$

$m \angle B+ 50 ^{\circ } = 90^{\circ }$

$m \angle B+ 50 ^{\circ }- 50 ^{\circ } = 90^{\circ } - 50 ^{\circ }$

$m \angle B = 40 ^{\circ }$

$m \angle A > m \angle B$ .

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ISEE Upper Level Quantitative : How to find the angle of a sector

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

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

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

Example Question #191 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #21 : Sectors

Example Question #191 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #191 : Plane Geometry

Example Question #17 : How To Find The Angle Of A Sector

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

All ISEE Upper Level Quantitative Resources

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