How to find the angle of a sector - ISEE Upper Level Mathematics Achievement

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Question

A giant clock has a minute hand four feet long. Since noon, the tip of the minute hand has traveled $10 \pi$ feet. What time is it now?

Answer

The circumference of the path traveled by the tip of the minute hand over the course of one hour is:

$C = 2 \pi r = 2 \pi \cdot 4 = 8 \pi$ feet.

Since the tip of the minute hand has traveled $10 \pi$ feet since noon, the minute hand has made

$\frac{10 \pi }{8 \pi } = \frac{5}{4} = 1 \frac{1}{4}$ revolutions. Therefore, $1 \frac{1}{4}$ hours have elapsed since noon, making the time 1:15 PM.

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Question

Note: Figure NOT drawn to scale

Refer to the above diagram. $\overarc {ABC}$ is a semicircle. Evaluate $m \overarc {AB}$ .

Answer

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, $\angle B$ , which intercepts the semicircle $\overarc{ABC}$ , is such an angle. Consequently,

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

$4x \div 4 = 108 \div 4$

$m \angle C = (x+10)^{\circ } = (27+10)^{\circ }= 37^{\circ }$

Inscribed $\angle C$ intercepts an arc with twice its angle measure; this arc is $\overarc{AB}$ , so

$m \widehat{AB} = 2 \cdot m \angle C = 2 \cdot 37 ^{\circ } = 74 ^{\circ }$ .

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Question

Figure NOT drawn to scale

Refer to the above diagram. $\overarc {ABC}$ is a semicircle. Evaluate $m \overarc {BCA}$ given $\measuredangle B=90^\circ$ .

Answer

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, $\angle B$ , which intercepts the semicircle $\overarc {ABC}$ , is such an angle. Consequently, $\bigtriangleup ABC$ is a right triangle, and $\angle A$ and $\angle C$ are complementary angles. Therefore,

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

$y + \left ( \frac{y}{3}+2\right ) = 90$

$y + \frac{1}{3}y +2 = 90$

$\frac{4}{3}y +2 = 90$

$\frac{4}{3}y +2 - 2 = 90 - 2$

$\frac{4}{3}y = 88$

$\frac{3}{4} \cdot \frac{4}{3}y = \frac{3}{4} \cdot 88$

$m \angle C = \left (\frac{y}{3}+ 2 \right ) ^{\circ }=\left ( \frac{66}{3}+ 2 \right )^{\circ }=( 22 + 2 )^{\circ }= 24^{\circ }$

Inscribed $\angle C$ intercepts an arc with twice its angle measure; this arc is $\overarc {AB}$ , so

$m \overarc {AB} = 2 \cdot m \angle C = 2 \cdot 24 ^{\circ } = 48 ^{\circ }$ .

The major arc corresponding to this minor arc, $\overarc {BCA}$ , has measure

$m \overarc {BCA} = 360 ^{\circ } - m \overarc {AB} = 360 ^{\circ } - 48^{\circ } = 312^{\circ }$

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Question

In the above diagram, radius .

Calculate the length of $\overarc {AB}$ .

Answer

Inscribed $\angle ACB$ , which measures $72 ^{\circ }$ , intercepts an arc with twice its measure. That arc is $\overarc {AB}$ , which consequently has measure

$72 ^{\circ } \times 2 = 144 ^{\circ }$ .

This makes $\overarc {AB}$ an arc which comprises

$\frac{144}{360} = \frac{144 \div 72}{360 \div 72} = \frac{2}{5}$

of the circle.

The circumference of a circle is $2 \pi$ multiplied by its radius, so

$C = 2 \pi r = 2\pi \cdot 20 = 40 \pi$ .

The length of $\overarc {AB}$ is $\frac{2}{5}$ of this, or $\frac{2}{5} \cdot 40 \pi = 16 \pi$ .

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Question

Figure NOT drawn to scale.

The circumference of the above circle is 100. $\overarc{AB }$ and $\overarc{CD}$ have lengths 20 and 15, respectively. Evaluate .

Answer

The length of $\overarc{AB }$ comprises $\frac{20}{100} = \frac{1}{5}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{5} \cdot 360 ^{\circ } = 72 ^{\circ }$ . Similarly, The length of $\overarc{CD}$ comprises $\frac{15}{100} = \frac{3}{20}$ of the circumference of the circle. Therefore, its degree measure is $\frac{3}{20} \cdot 360 ^{\circ } = 54 ^{\circ }$ .

If two chords cut each other inside the circle, as $\overline{AC }$ and $\overline{BD}$ do, and one pair of vertical angles are examined, then the degree measure of each angle is half the sum of those of the arcs intercepted - that is,

$t = \frac{1}{2} ( m \overarc{AB}+ m \overarc {CD})$

$t = \frac{1}{2} ( 72 ^{\circ } + 54 ^{\circ })$

$t = \frac{1}{2} ( 126^{\circ })$

$t =63 ^{\circ }$

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Question

Figure NOT drawn to scale.

The circumference of the above circle is 120. $\overarc{AB }$ and $\overarc{CD}$ have lengths 10 and 20, respectively. Evaluate .

Answer

The length of $\overarc{AB }$ comprises $\frac{10}{120} = \frac{1}{12}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{12} \cdot 360 ^{\circ } = 30^{\circ }$ . Similarly, The length of $\overarc{CD}$ comprises $\frac{20}{120} = \frac{1}{6}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{6} \cdot 360 ^{\circ } = 60 ^{\circ }$ .

If two secants are constructed to a circle from an outside point, the degree measure of the angle the secants form is half the difference of those of the arcs intercepted - that is,

$t = \frac{1}{2} ( m \overarc {CD} - m \overarc{AB})$

$t = \frac{1}{2} ( 60 - 30 ) = \frac{1}{2} \cdot 30 = 15$ .

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Question

Figure NOT drawn to scale.

Refer to the above diagram. $\overarc{AB }$ and $\overarc{ABC}$ have lengths 80 and 160, respectively. Evaluate .

Answer

The circumference of the circle is the sum of the two arc lengths:

The length of $\overarc{AB }$ comprises $\frac{80}{240} = \frac{1}{3}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{3} \cdot 360 ^{\circ } = 120 ^{\circ }$ . Consequently, $\overarc{ACB }$ is an arc of degree measure $360 ^{\circ } - 120 ^{\circ } = 240^{\circ }$ .

The segments shown are both tangents from to the circle. Consequently, the degree measure of the angle they form is half the difference of the angle measures of the arcs they intercept - that is,

$t = \frac{1}{2} ( m \overarc {ACB} - m \overarc{AB})$

$t = \frac{1}{2} ( 240 ^{\circ } - 120^{\circ }) = \frac{1}{2} \cdot 120^{\circ } = 60 ^{\circ }$

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