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ISEE Upper Level Quantitative : Sectors

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Example Questions

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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 : Sectors

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

All ISEE Upper Level Quantitative Resources

Example Questions

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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