Circle Theorems
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Length of Arc, Area of Sector, Area of Circle, Circumference of Circle
Except stated otherwise, use:
$
d = diameter \\[3ex]
r = radius \\[3ex]
L = arc\:\:length \\[3ex]
A = area\;\;of\;\;sector \\[3ex]
\theta = central\:\:angle \\[3ex]
\pi = \dfrac{22}{7} \\[5ex]
RAD = radians \\[3ex]
^\circ = DEG = degrees \\[7ex]
\underline{\theta\;\;in\;\;DEG} \\[3ex]
L = \dfrac{2\pi r\theta}{360} \\[5ex]
\theta = \dfrac{180L}{\pi r} \\[5ex]
r = \dfrac{180L}{\pi \theta} \\[5ex]
A = \dfrac{\pi r^2\theta}{360} \\[5ex]
\theta = \dfrac{360A}{\pi r^2} \\[5ex]
r = \dfrac{360A}{\pi\theta} \\[7ex]
\underline{\theta\;\;in\;\;RAD} \\[3ex]
L = r\theta \\[5ex]
\theta = \dfrac{L}{r} \\[5ex]
r = \dfrac{L}{\theta} \\[5ex]
A = \dfrac{r^2\theta}{2} \\[5ex]
\theta = \dfrac{2A}{r^2} \\[5ex]
r = \sqrt{\dfrac{2A}{\theta}} \\[7ex]
Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex]
L = \dfrac{2A}{r} \\[5ex]
r = \dfrac{2A}{L} \\[5ex]
A = \dfrac{Lr}{2}
$
Circle Theorems
(1.) The angle in a semicircle is a right angle (an angle of 90°).
(2.) Angles in the same segment of a circle are equal.
OR
Angles subtended by a chord of a circle in the same segment of the circle are equal.
(3.) The angle which an arc of a circle subtends at the center is twice the angle which the same
arc of the circle subtends at the circumference.
OR
The measure of any angle inscribed in a circle is half the measure of the intercepted arc.
(4.) The sum of the interior opposite angles of a cyclic quadrilateral is 180°
OR
The interior opposite angles of a cyclic quadrilateral are supplementary
(5.) The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
(6.) The radius of a circle is perpendicular to the tangent of the circle at the point of contact.
This implies that the angle between the radius of a circle and the tangent to the circle at the point of
contact is 90°
(7.) Intersecting Tangents Theorem or Intersecting Tangent-Tangent Theorem and
Angle of Intersecting Tangents Theorem
If two tangents are drawn from the same external point:
(a.) the two tangents are equal in length
(b.) the line joining the external point and the centre of the circle bisects the angle formed by the two
tangents.
(c.) the line joining the external point and the centre of the circle bisects the angle formed by the two
radii.
(8.) Alternate Segment Theorem
The angle between a tangent to a circle and a chord drawn from the point of contact, is equal to the angle in
the alternate segment.
(9.) If a line drawn from the center of the circle bisects a chord, then:
(a.) it bisects its arc (the angle opposite the chord) and
(b.) it is perpendicular to the chord.
(10.) If a line drawn from the center of the circle is perpendicular to a chord, then:
(a.) it bisects the chord and
(b.) it bisects its arc (the angle opposite the chord).
(11.) Intersecting Chords Theorem
When two chords intersect, the product of the lengths of the segments of one chord is equal to the product of
the
lengths of the segments of the other chord.
(12.) Angle of Intersecting Chords Theorem
The angle formed when two chords intersect is equal to half the sum of the intercepted arcs.
(13.) Intersecting Secants Theorem or Intersecting Secant-Secant Theorem
In the intersection of two secants from the same external point:
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the first secant is equal to
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the second secant.
(14.) Angle of Intersecting Secants (Inside the Circle) Theorem
The angle formed when two secants intersect inside a circle is equal to half the sum of the intercepted arcs.
(15.) Angle of Intersecting Secants (Outside the Circle) Theorem
The angle formed when two secants intersect outside a circle is equal to half the difference of the
intercepted arcs.
(16.) Intersecting Secant-Tangent Theorem or Intersecting Tangent-Secant Theorem
In the intersection of a secant and a tangent from the same external point:
the product of the distance between the first point and the external point and the distance between the
second point and the external point for the secant is equal to
the square of the distance between the point of contant and the external point for the tangent.
(17.) Angle of Intersecting Secant-Tangent Theorem
The angle formed when a secant and a tangent intersect outside a circle is equal to half the difference of the
intercepted arcs.
The length of $\overline{AO}$ is 4 inches, and arc $\overset{\huge\frown}{AB}$ measures 112°
What is the measure of ∠ABO?
$ F.\;\; 17^\circ \\[3ex] G.\;\; 34^\circ \\[3ex] H.\;\; 45^\circ \\[3ex] J.\;\; 56^\circ \\[3ex] K.\;\; 68^\circ \\[3ex] $
Intercepted arc = angle at center (central angle)
$ \angle AOB = \overset{\huge\frown}{AB} = 112^\circ \\[5ex] \underline{\triangle AOB} \\[3ex] \angle ABO = \angle BAO = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] \angle ABO + \angle BAO + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAB \\[3ex] \implies \\[3ex] p + p + 112 = 180 \\[3ex] 2p = 180 - 112 \\[3ex] 2p = 68 \\[3ex] p = \dfrac{68}{2} \\[5ex] p = 34 \\[3ex] \therefore \angle ABO = \angle BAO = 34^\circ $
What is the degree measure of $\angle ABC$?
$ A.\;\; 45^\circ \\[3ex] B.\;\; 60^\circ \\[3ex] C.\;\; 75^\circ \\[3ex] D.\;\; 90^\circ \\[3ex] E.\;\; \text{Cannot be determined from the given information} \\[3ex] $
$ \angle ABC = 90^\circ...\text{Angle in a semicircle} $
What is the sum of the measures of $\angle CMO$ and $\angle CNO$?
$ F.\;\; 90^\circ \\[3ex] G.\;\; 67.5^\circ \\[3ex] H.\;\; 60^\circ \\[3ex] J.\;\; 45^\circ \\[3ex] K.\;\; 22.5^\circ \\[3ex] $
$ Center:\;\;O \\[3ex] \angle \;\;at\;\;center = 90^\circ \\[3ex] Let: \\[3ex] \angle MCN = \psi \\[3ex] \angle CMN = \theta \\[3ex] \angle CNM = \phi \\[3ex] \angle OMN = x \\[3ex] \angle ONM = y \\[3ex] 90 = 2 * \psi ...\angle \;\;at\;\;center = 2 * \angle\;\;at\;\;circumference \\[3ex] 2\psi = 90 \\[3ex] \psi = \dfrac{90}{2} \\[5ex] \psi = 45^\circ \\[3ex] Construction:\;\;Draw\;\;a\;\;chord\;\;to\;\;join\;\;M\;\;to\;\;N \\[5ex] \underline{\triangle OMN} \\[3ex] x = y ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OMN \\[3ex] Because\;\;of\;\;same\;\;radii...OM = ON \\[3ex] \implies \\[3ex] 90 + x + y = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OMN \\[3ex] 90 + x + x = 180 \\[3ex] 90 + 2x = 180 \\[3ex] 2x = 180 - 90 \\[3ex] 2x = 90 \\[3ex] x = \dfrac{90}{2} \\[5ex] x = 45 \\[3ex] x = y = 45^\circ \\[5ex] \underline{\triangle CMN} \\[3ex] \psi + \theta + x + y + \phi = 180 ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CMN \\[3ex] 45 + \theta + 45 + 45 + \phi = 180 \\[3ex] 135 + \theta + \phi = 180 \\[3ex] \theta + \phi = 180 - 135 \\[3ex] \theta + \phi = 45^\circ $
Points W, X, Y, and Z lie on the circle.
The measure of $\angle WCY$ is $100^\circ$, the measure of $\angle XCZ$ is $80^\circ$, and the measure of $\angle WCZ$ is $150^\circ$
What is the measure of $\angle XCY$?
$ A.\;\; 20^\circ \\[3ex] B.\;\; 25^\circ \\[3ex] C.\;\; 30^\circ \\[3ex] D.\;\; 40^\circ \\[3ex] E.\;\; 50^\circ \\[3ex] $
The ACT is a timed test.
It is important to do this question within a minute. Remember what I wrote earlier
So, the faster approach for me is to use alphabets rather than angles
This is what I mean
$ Let: \\[3ex] \angle WCX = m \\[3ex] \angle XCY = n \\[3ex] \angle YCZ = p \\[3ex] Based\;\;on\;\;the\;\;question\;\;and\;\;modified\;\;diagram: \\[3ex] m + n = 100 ...eqn.(1) \\[3ex] n + p = 80...eqn.(2) \\[3ex] m + n + p = 150 ...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;into\;\;eqn.(3) \implies \\[3ex] 100 + p = 150 \\[3ex] p = 150 - 100 \\[3ex] p = 50 ...eqn.(4) \\[3ex] Substitute\;\;eqn.(4)\;\;into\;\;eqn.(2) \implies \\[3ex] n + 50 = 80 \\[3ex] n = 80 - 50 \\[3ex] n = 30 \\[3ex] \therefore \angle XCY = 30^\circ $
What is the measure of $\angle ABC$?
$ F.\;\; 35^\circ \\[3ex] G.\;\; 40^\circ \\[3ex] H.\;\; 45^\circ \\[3ex] J.\;\; 55^\circ \\[3ex] K.\;\; \text{Cannot be determined from the given information} \\[3ex] $
$ \underline{\triangle CAB} \\[3ex] \angle BAC = \angle ABC = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle CAB \\[3ex] \angle BAC + \angle ABC + \angle BCA = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CAB \\[3ex] p + p + 110 = 180 \\[3ex] 2p + 110 = 180 \\[3ex] 2p = 180 - 110 \\[3ex] 2p = 70 \\[3ex] p = \dfrac{70}{2} \\[5ex] p = 35 \\[3ex] \therefore \angle ABC = 35^\circ $
Which of the following does NOT appear in the figure?
A. Acute triangle
B. Equilateral triangle
C. Isosceles triangle
D. Right triangle
E. Scalene triangle
Let us analyze the options
$ \angle AOB = 60^\circ...implies\;\;an\;\;acute\;\;\triangle \\[3ex] $ An acute triangle is a triangle that has three angles less than $90^\circ$
Option A appears in the figure
$ \underline{\triangle AOB} \\[3ex] OA = OB ...same\;\;radii \\[3ex] \implies \triangle AOB\;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] \angle OBA = \angle OAB ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle AOB \\[3ex] $ Option C appears in the figure
$ \angle OBA + \angle OAB + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] Let\;\;\angle OBA = \angle OAB = p \\[3ex] \angle AOB = 60^\circ \\[5ex] \underline{\triangle AOB} \\[3ex] p + p + 60 = 180 \\[3ex] 2p = 180 - 60 \\[3ex] 2p = 120 \\[3ex] p = \dfrac{120}{2} \\[5ex] p = 60 \\[3ex] \therefore \angle OBA = \angle OAB = 60^\circ \\[3ex] $ Each angle in $\triangle AOB = 60^\circ$
This implies that $\triangle AOB$ is an equilateral triangle.
Option B appears in the figure '
$ \underline{\triangle COD} \\[3ex] \angle COD = 90^\circ ...implies\;\;a\;\;right\;\;\triangle \\[3ex] $ A right triangle is a triangle that has one angle of $90^\circ$ and two angles less than $90^\circ$ Option D appears in the figure
The only remaining option is Option E
Points A and C are on the circle.
The measure of $\angle ABC$ is $95^\circ$ and the measure of $\angle BCD$ is $85^\circ$
The lines in which of the following pairs of lines are necessarily parallel?
$ I.\;\; l\;\;and\;\;m \\[3ex] II.\;\; \overleftrightarrow{AB} \;\;and\;\; \overleftrightarrow{DC} \\[3ex] III.\;\; \overleftrightarrow{AD} \;\;and\;\; \overleftrightarrow{BC} \\[3ex] F.\;\; I\;\;only \\[3ex] G.\;\; II\;\;only \\[3ex] H.\;\; III\;\;only \\[3ex] J.\;\; I\;\;and\;\;II\;\;only \\[3ex] K.\;\; I,\;\;II,\;\;and\;\;III \\[3ex] $
$ l\;\;and\;\;m\;\;are\;\;not\;\;parallel \\[3ex] Eliminate\;\;Option\;I \\[5ex] \underline{Cyclic\;\;Quadrilateral\;ABCD} \\[3ex] \angle A + \angle C = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle A + 85 = 180 \\[3ex] \angle A = 180 - 85 \\[3ex] \angle A = 95^\circ \\[3ex] Also: \\[3ex] \angle B + \angle D = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 95 + \angle D = 180 \\[3ex] \angle D = 180 - 95 \\[3ex] \angle D = 85^\circ \\[3ex] Try\;\;Option\;II \\[3ex] $
$ If\;\;\overleftrightarrow{AB} \;||\; \overleftrightarrow{DC} \\[3ex] e = 85^\circ ...corresponding\;\;\angle s\;\;are\;\;congruent \\[3ex] But: \\[3ex] e + 95^\circ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] e = 180 - 95 \\[3ex] e = 85^\circ \\[3ex] 85^\circ = 85^\circ \\[3ex] \implies that\;\; \overleftrightarrow{AB} \;||\; \overleftrightarrow{DC} \\[3ex] Try\;\;Option\;III \\[3ex] $
$ If\;\;\overleftrightarrow{AD} \;||\; \overleftrightarrow{BC} \\[3ex] f = 95^\circ ...corresponding\;\;\angle s\;\;are\;\;congruent \\[3ex] But: \\[3ex] f + 95^\circ = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] f = 180 - 95 \\[3ex] f = 85^\circ \\[3ex] 95^\circ \ne 85^\circ \\[3ex] \implies that\;\; \overleftrightarrow{AD} \;\not||\; \overleftrightarrow{BC} \\[3ex] Eliminate\;\;Option\;III \\[3ex] Correct\;\;Option:\;\;II\;\;only $
What is the value of x?
(Note: When two chords intersect, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.)
$ A.\;\; 10 \\[3ex] B.\;\; 13.5 \\[3ex] C.\;\; 14 \\[3ex] D.\;\; 17.5 \\[3ex] E.\;\; 19 \\[3ex] $
$ 2 * x = 7 * 5 ...Intersecting\;\;Chords\;\;Theorem \\[3ex] x = \dfrac{7 * 5}{2} \\[5ex] x = \dfrac{35}{2} \\[5ex] x = 17.5 $
What is the measure of $\angle BAO$?
$ F.\;\; 30^\circ \\[3ex] G.\;\; 40^\circ \\[3ex] H.\;\; 50^\circ \\[3ex] J.\;\; 60^\circ \\[3ex] K.\;\; 80^\circ \\[3ex] $
Construction: Join the radius from point O to point B
Intercepted arc = angle at center (central angle)
$ \angle AOB = \overset{\huge\frown}{AB} = 80^\circ \\[5ex] \underline{\triangle OAB} \\[3ex] \angle BAO = \angle ABO = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle OAB \\[3ex] \angle BAO + \angle ABO + \angle AOB = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OAB \\[3ex] \implies \\[3ex] p + p + 80 = 180 \\[3ex] 2p = 180 - 80 \\[3ex] 2p = 100 \\[3ex] p = \dfrac{100}{2} \\[5ex] p = 50 \\[3ex] \therefore \angle BAO = \angle ABO = 50^\circ $
Which of the following is the most direct explanation of why $\triangle MCA$ is isosceles?
F. 2 sides are radii of the circle
G. Side-angle-side congruence
H. Angle-side-angle congruence
J. Angle-angle-angle similarity
K. The Pythagorean theorem
Construction: Draw the radius from center: M to circumference: C
Because MC and MA are radii (same length):
$\angle MCA = \angle MAC$...base angles of isosceles $\triangle MCA$
$\overline{AE}$ and $\overline{BD}$ both go through C, q is tangent to the circle at B, F lies on $\overleftrightarrow{AD}$, and the measure of $\angle FAC$ is $145^\circ$
What is the measure of $\angle BCE$?
$ F.\;\; 145^\circ \\[3ex] G.\;\; 90^\circ \\[3ex] H.\;\; 65^\circ \\[3ex] J.\;\; 55^\circ \\[3ex] K.\;\; 35^\circ \\[3ex] $
$ \theta = 145^\circ...interior\;\;alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BEC + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle BEC + 145 = 180 \\[3ex] \angle BEC = 180 - 145 \\[3ex] \angle BEC = 35^\circ \\[3ex] \angle CBE = 90^\circ \\[3ex] ...radius\;CB\;\perp\;tangent\;qBE\;\;at\;\;point\;\;of\;\;contact\;B \\[5ex] \underline{\triangle CBE} \\[3ex] \angle BCE + \angle BEC + \angle CBE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle CBE \\[3ex] \angle BCE + 35 + 90 = 180 \\[3ex] \angle BCE + 125 = 180 \\[3ex] \angle BCE = 180 - 125 \\[3ex] \angle BCE = 55^\circ $
If O marks the center of the circle, what is the length of $\overline{OC}$ ?
$ A.\;\; 2\sqrt{3} \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 12 \\[3ex] D.\;\; 4\sqrt{21} \\[3ex] E.\;\; 36 \\[3ex] $
Construction: Draw the radius from Point O to POint A
$ |AC| = |CB| = \dfrac{|AB|}{2} \\[5ex] ...line\;OA\;\;\perp\;\;to\;\;chord\;AB;\;\;bisects\;\;the\;\;chord \\[3ex] |AC| = \dfrac{16}{2} = 8\;cm \\[5ex] |OA| = 10\;cm ...radius \\[5ex] \underline{Right\;\triangle\;\;OCA} \\[3ex] |OC|^2 + |AC|^2 = |OA|^2 ...Pythagorean\;\;theorem \\[3ex] |OC|^2 + 8^2 = 10^2 \\[3ex] |OC|^2 + 64 = 100 \\[3ex] |OC|^2 = 100 - 64 \\[3ex] |OC|^2 = 36 \\[3ex] |OC| = \sqrt{36} \\[3ex] |OC| = 6\;cm $
Which of the following statements is NOT true?
$ F.\;\; \angle TUM \;\;measures\;\;65^\circ \\[3ex] G.\;\; \overline{TU}\;\;is\;\;parallel\;\;to\;\;\overline{RS} \\[3ex] H.\;\; \overset{\Huge\frown}{TXU}\;\;measures\;\;50^\circ \\[3ex] J.\;\; \overline{RM} \cong \overline{TM} \\[3ex] K.\;\; \overline{RS} \cong \overline{SM} \\[3ex] $
Let us analyze the options
$ \angle TMU = 50^\circ ...vertical \angle s\;\;are\;\;equal \\[5ex] \underline{\triangle MTU} \\[3ex] \angle UTM = \angle TUM = p ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MTU \\[3ex] \angle UTM + \angle TUM + \angle TMU = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MTU \\[3ex] p + p + 50 = 180 \\[3ex] 2p + 50 = 180 \\[3ex] 2p = 180 - 50 \\[3ex] 2p = 130 \\[3ex] p = \dfrac{130}{2} \\[5ex] p = 65 \\[3ex] \therefore \angle UTM = \angle TUM = 65^\circ \\[3ex] Option\;\;F\;\;is\;\;true \\[5ex] \underline{\triangle MRS} \\[3ex] \angle SRM = \angle RSM = k ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MRS \\[3ex] \angle SRM + \angle RSM + \angle RMS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle MRS \\[3ex] k + k + 50 = 180 \\[3ex] 2k + 50 = 180 \\[3ex] 2k = 180 - 50 \\[3ex] 2k = 130 \\[3ex] k = \dfrac{130}{2} \\[5ex] k = 65 \\[3ex] \therefore \angle SRM = \angle RSM = 65^\circ \\[3ex] \angle SRM = \angle SRU = 65^\circ ...diagram \\[3ex] \angle TUM = \angle TUR = 65^\circ ...diagram \\[3ex] Because: \\[3ex] \angle SRU = \angle TUR = 65^\circ ...alternate \angle s\;\;are\;\;equal \\[3ex] \implies \;\;that\;\; \overline{RS} \;\;||\;\; \overline{TU} \\[3ex] Option\;\;G\;\;is\;\;true \\[3ex] \overset{\Huge\frown}{TXU} = 50^\circ ...vertical \angle s\;\;are\;\;equal \\[3ex] ...also\;\;\angle\;\;subtended\;\;by\;\;arc\;TXU \\[3ex] Option\;\;H\;\;is\;\;true \\[3ex] \overline{RM} = \overline{TM} ...same\;\;radii\;\;from\;\;center\;M \\[3ex] Option\;\;J\;\;is\;\;true \\[3ex] For\;\;the\;\;remaining\;\; Option\;\;K \\[3ex] \overline{RM} = \overline{SM} ...same\;\;radii\;\;from\;\;center\;M \\[3ex] \implies\;\; \triangle RMS\;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] However: \\[3ex] \overline{RS} \ncong \overline{SM} \\[3ex] $ Because $\triangle RMS$ is not an equilateral triangle
An equilateral triangle must have each of the three angles as $60^\circ$ in order for all sides to be congruent
$ Option\;\;K\;\;is\;\;NOT\;\;true $
The circle intersects $\overline{OP}$ at B.
The length of $\overline{AP}$ is 12 centimeters, and the measure of ∠APO is 20°.
Which of the following values is closest to the length, in centimeters, of $\overline{BP}$ ?
(Note: $\sin 20^\circ \approx 0.342$, $\cos 20^\circ \approx 0.940$, and $\tan 20^\circ \approx 0.364$)
$ A.\;\; 2.1 \\[3ex] B.\;\; 4.4 \\[3ex] C.\;\; 6.9 \\[3ex] D.\;\; 7.6 \\[3ex] E.\;\; 8.4 \\[3ex] $
$ SOHCAHTOA \\[3ex] \tan 20 = \dfrac{\overline{OA}}{\overline{AP}} \\[5ex] 0.364 = \dfrac{\overline{OA}}{12} \\[5ex] \overline{OA} = 12 * 0.364 \\[3ex] \overline{OA} = 4.368 \\[3ex] \overline{OB} = \overline{OA} = 4.368\;cm ...radius \\[3ex] \cos 20 = \dfrac{\overline{AP}}{\overline{OP}} \\[5ex] 0.940 = \dfrac{12}{\overline{OP}} \\[5ex] 0.94 * \overline{OP} = 12 \\[3ex] \overline{OP} = \dfrac{12}{0.94} \\[5ex] \overline{OP} = 12.76595745 \\[3ex] \overline{OP} = \overline{OB} + \overline{BP} ...diagram \\[3ex] \overline{OB} + \overline{BP} = \overline{OP} \\[3ex] 4.368 + \overline{BP} = 12.76595745 \\[3ex] \overline{BP} = 12.76595745 - 4.368 \\[3ex] \overline{BP} = 8.397957447 \\[3ex] \overline{BP} \approx 8.4\;cm $
What is the degree measure of minor arc $\overset{\huge\frown}{RS}$?
$ F.\;\; 30^\circ \\[3ex] G.\;\; 45^\circ \\[3ex] H.\;\; 60^\circ \\[3ex] J.\;\; 90^\circ \\[3ex] K.\;\; \text{Cannot be determined from the given information} \\[3ex] $
$ First\;\;Approach:\;\;Recommmended\;\;for\;\;ACT \\[5ex] \underline{\triangle PST} \\[3ex] \angle PST = \angle PTS = 30^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle PST \\[3ex] \overset{\huge\frown}{RS} = \angle PST + \angle PTS \\[3ex] ...exterior\;\;\angle\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s\\[3ex] \overset{\huge\frown}{RS} = 30 + 30 \\[3ex] \overset{\huge\frown}{RS} = 60^\circ \\[3ex] OR \\[3ex] Second\;\;Approach: \\[3ex] \angle PST + \angle PTS + \angle TPS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle PST \\[3ex] 30 + 30 + \angle TPS = 180 \\[3ex] 60 + \angle TPS = 180 \\[3ex] \angle TPS = 180 - 60 \\[3ex] \angle TPS = 120^\circ \\[3ex] \angle TPS + \overset{\huge\frown}{RS} = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 120 + \overset{\huge\frown}{RS} = 180 \\[3ex] \overset{\huge\frown}{RS} = 180 - 120 \\[3ex] \overset{\huge\frown}{RS} = 60^\circ $
Line l passes through O(0, 0) and is tangent to the circle at A, and B is the center of the circle.
What is the measure of $\angle AOB$?
$ F.\;\; 15^\circ \\[3ex] G.\;\; 22.5^\circ \\[3ex] H.\;\; 30^\circ \\[3ex] J.\;\; 45^\circ \\[3ex] K.\;\; 60^\circ \\[3ex] $
$ \underline{Equation\;\;of\;a\;\;circle\;\;with\;\;center\;(h, k)} \\[3ex] (x - k)^2 + (y - k)^2 = r^2 \\[3ex] Compare\;\;with \\[3ex] (x - 2)^2 + y^2 = 1 \\[3ex] \implies \\[3ex] (x - 2)^2 + (y - 0)^2 = 1^2 \\[3ex] center = (h, k) = (2, 0) \\[3ex] radius = r = 1\;unit \\[3ex] r = |AB| = 1\;unit ...diagram \\[3ex] |OB| = distance\;\;between\;\;O(0, 0)\;\;and\;\;B(2, 0) \\[3ex] |OB| = \sqrt{(2 - 0)^2 + (0 - 0)^2}... Distance\;\;Formula \\[3ex] |OB| = \sqrt{2^2 + 0} \\[3ex] |OB| = \sqrt{4} \\[3ex] |OB| = 2\;units \\[3ex] \dfrac{\sin \angle AOB}{|AB|} = \dfrac{\sin \angle OAB}{|OB|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle AOB}{1} = \dfrac{\sin 90}{2} \\[5ex] \sin \angle AOB = \dfrac{1}{2} \\[5ex] \angle AOB = \sin^{-1}\left(\dfrac{1}{2}\right) \\[5ex] \angle AOB = 30^\circ $
Points A, B, and C are on the circle shown below.
What is the measure of ∠ABC?
$ A.\;\; 40^\circ \\[3ex] B.\;\; 60^\circ \\[3ex] C.\;\; 80^\circ \\[3ex] D.\;\; 140^\circ \\[3ex] E.\;\; 160^\circ \\[3ex] $
Construction: Join the chord from point A to point C
$ 190 = 2 * \angle BAC ...intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] 2 * \angle BAC = 190 \\[3ex] \angle BAC = \dfrac{190}{2} \\[5ex] \angle BAC = 95^\circ \\[3ex] 90 = 2 * \angle BCA ...intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] 2 * \angle BCA = 90 \\[3ex] \angle BCA = \dfrac{90}{2} \\[5ex] \angle BCA = 45^\circ \\[5ex] \underline{\triangle BAC} \\[3ex] \angle ABC + \angle BAC + \angle BCA = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle BAC \\[3ex] \angle ABC + 95 + 45 = 180 \\[3ex] \angle ABC + 140 = 180 \\[3ex] \angle ABC = 180 - 140 \\[3ex] \angle ABC = 40^\circ $
The measure of $\angle CAD$ is $30^\circ$, and the measure of minor arc $\overset{\huge\frown}{CD}$ is $60^\circ$
What is the measure of minor arc $\overset{\huge\frown}{AC}$
$ F.\;\; 75^\circ \\[3ex] G.\;\; 90^\circ \\[3ex] H.\;\; 105^\circ \\[3ex] J.\;\; 120^\circ \\[3ex] K.\;\; 150^\circ \\[3ex] $
$ \overline{AD}\;\;is\;\;a\;\;diameter \\[3ex] \angle ACD = 90^\circ...\angle \;\;in\;\;a\;\;semicircle \\[5ex] \underline{\triangle ACD} \\[3ex] \angle ADC + \angle ACD + \angle CAD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] \angle ADC + 90 + 30 = 180 \\[3ex] \angle ADC + 120 = 180 \\[3ex] \angle ADC = 180 - 120 \\[3ex] \angle ADC = 60^\circ \\[3ex] \overset{\huge\frown}{AC} = 2 * \angle ADC ... intercepted\;\;arc = 2 * inscribed\;\;\angle \\[3ex] \overset{\huge\frown}{AC} = 2 * 60 \\[3ex] \overset{\huge\frown}{AC} = 120^\circ $
Segment $\overline{BC}$ is then drawn to form $\triangle ABC$
If $\angle A$ measures $70^\circ$, what is the measure of $\angle ABC$?
$ F.\;\; 70^\circ \\[3ex] G.\;\; 55^\circ \\[3ex] H.\;\; 40^\circ \\[3ex] J.\;\; 35^\circ \\[3ex] K.\;\; \text{Cannot be determined from the given information} \\[3ex] $
Let us draw the diagram for this question
AB = AC ...the two tangents: AB and AC, drawn from the same external point:A are equal in length
$ \implies \triangle ABC\;\;is\;\;an\;\;isosceles\;\;triangle \\[5ex] \underline{\triangle ABC} \\[3ex] \angle ABC + \angle ACB + \angle BAC = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] Let\;\; \angle ABC = \angle ACB = p \\[3ex] \angle BAC = \angle A = 70^\circ \\[3ex] \implies \\[3ex] p + p + 70 = 180 \\[3ex] 2p = 180 - 70 \\[3ex] 2p = 110 \\[3ex] p = \dfrac{110}{2} \\[5ex] p = 55 \\[3ex] \therefore \angle ABC = 55^\circ $
What is the measure of ∠CAD?
$ F.\;\; 27^\circ \\[3ex] G.\;\; 30^\circ \\[3ex] H.\;\; 37.5^\circ \\[3ex] J.\;\; 40^\circ \\[3ex] K.\;\; 60^\circ \\[3ex] $
$ |AD|\;\;is\;\;a\;\;diameter \\[3ex] \implies \\[3ex] \angle ACD = 90^\circ ...\angle\;\;in\;\;a\;\;semicircle \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;hexagon:\;\;n = 6 \;\;(6\;\;sides) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = 180(6 - 2) \\[3ex] = 180(4) \\[3ex] = 720^\circ \\[3ex] Each\;\;interior\;\;\angle\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = \angle F = 120^\circ \\[5ex] \underline{Cyclic\;\;Quadrilateral\;\;ABCD} \\[3ex] \angle ABC = \angle B = 120^\circ ...diagram \\[3ex] \angle ABC + \angle CDA = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] 120 + \angle CDA = 180 \\[3ex] \angle CDA = 180 - 120 \\[3ex] \angle CDA = 60^\circ \\[5ex] \underline{\triangle ACD} \\[3ex] \angle CAD + \angle CDA + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACD \\[3ex] \angle CAD + 60 + 90 = 180 \\[3ex] \angle CAD + 150 = 180 \\[3ex] \angle CAD = 180 - 150 \\[3ex] \angle CAD = 30^\circ $
If the length of radius $\overline{OD}$ is 15 centimeters, how long is $\overline{AB}$, in centimeters?
$ A.\;\; 15 \\[3ex] B.\;\; 18 \\[3ex] C.\;\; 30 \\[3ex] D.\;\; 5\pi \\[3ex] E.\;\; \dfrac{225\pi}{6} \\[5ex] $
Construction: Join the chord from Point B to Point E
$ Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;polygon \\[3ex] = 180(n - 2) \\[3ex] For\;\;a\;\;regular\;\;hexagon:\;\;n = 6 \;\;(6\;\;sides) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = 180(6 - 2) \\[3ex] = 180(4) \\[3ex] = 720^\circ \\[3ex] Each\;\;interior\;\;\angle\;\;of\;\;a\;\;regular\;\;hexagon \\[3ex] = \dfrac{Sum}{n} \\[5ex] = \dfrac{720}{6} \\[5ex] = 120^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = \angle F = 120^\circ \\[3ex] \underline{Cyclic\;\;Quadrilateral\;\;EBCD} \\[3ex] \angle BCD = \angle C = 120^\circ ...diagram \\[3ex] \angle BED + \angle BCD = 180^\circ \\[3ex] ...interior\;\;opposite\;\;\angle s\;\;of\;\;a\;\;cyclic\;\;quad\;\;are\;\;supplementary \\[3ex] \angle BED + 120 = 180 \\[3ex] \angle BED = 180 - 120 \\[3ex] \angle BED = 60^\circ \\[3ex] \underline{\triangle OED} \\[3ex] \angle OED = \angle BED = 60^\circ ... diagram \\[3ex] \angle OED = \angle ODE = 60^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle OED \\[3ex] \implies \\[3ex] |OE| = |OD| = 15\;cm ... isosceles\;\; \triangle OED \\[3ex] \angle EOD + \angle OED + \angle ODE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle OED \\[3ex] \angle EOD + 60 + 60 = 180 \\[3ex] \angle EOD + 120 = 180 \\[3ex] \angle EOD = 180 - 120 \\[3ex] \angle EOD = 60^\circ \\[3ex] \implies \\[3ex] \triangle OED \;\;is\;\;an\;\;equilateral\;\;\triangle \\[3ex] \angle EOD = \angle OED = \angle ODE = 60^\circ \\[3ex] Similarly: \\[3ex] |OD| = |OE| = |ED| = 15\;cm ...equilateral\;\; \triangle OED \\[3ex] \implies \\[3ex] |AB| = |ED| = 15\;cm ...regular\;\;hexagon $
Secants $\overleftrightarrow{AC}$ and $\overleftrightarrow{CE}$ intersect at C.
The chords $overline{AB}$ and $\overline{DE}$ are congruent.
Minor arc $\overset{\huge\frown}{AE}$ measures 106°
Minor arc $\overset{\huge\frown}{BD}$ measures 64°
What is the measure of minor arc $\overset{\huge\frown}{DE}$ ?
$ F.\;\; 42^\circ \\[3ex] G.\;\; 84^\circ \\[3ex] H.\;\; 85^\circ \\[3ex] J.\;\; 90^\circ \\[3ex] K.\;\; 95^\circ \\[3ex] $
Let us represent the information in the circle
Construction:
(a.) Let O be the center of the circle.
(b.) Indicate the central angles corresponding to the minor arcs.
(c.) Indicate the congruent lines (chords)
$ \underline{\triangle AOE} \\[3ex] \angle OAE = \angle OEA = p ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle AOE \\[3ex] \angle OAE + \angle OEA + \angle AOE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle AOE \\[3ex] p + p + 106 = 180 \\[3ex] 2p = 180 - 106 \\[3ex] 2p = 74 \\[3ex] p = \dfrac{74}{2} \\[5ex] p = 37 \\[3ex] \therefore \angle OAE = \angle OEA = 37^\circ \\[3ex] \angle ACE = \dfrac{\overset{\huge\frown}{AE} - \overset{\huge\frown}{BD}}{2} ... Angle\;\;of\;\;Intersecting\;\;Chords\;\;Theorem \\[5ex] \implies \\[3ex] \angle ACE = \dfrac{\angle AOE - \angle BOD}{2} \\[5ex] \angle ACE = \dfrac{106 - 64}{2} \\[5ex] \angle ACE = \dfrac{42}{2} \\[5ex] \angle ACE = 21^\circ \\[3ex] \underline{\triangle ACE} \\[3ex] |AB| \cong |DE| ...given \\[3ex] \implies \\[3ex] \angle CAE = \angle CEA = k \\[3ex] \angle CAE + \angle CEA + \angle ACE = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ACE \\[3ex] k + k + 21 = 180 \\[3ex] 2k = 180 - 21 \\[3ex] 2k = 159 \\[3ex] k = \dfrac{159}{2} \\[5ex] k = 79.5 \\[3ex] \therefore \angle CAE = \angle CEA = 79.5^\circ \\[3ex] \angle OEA + \angle OED = \angle CEA ...diagram \\[3ex] 37 + \angle OED = 79.5 \\[3ex] \angle OED = 79.5 - 37 \\[3ex] \angle OED = 42.5^\circ \\[3ex] \underline{\triangle EOD} \\[3ex] \angle OED = \angle ODE = 42.5^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\; \triangle EOD \\[3ex] \angle OED + \angle ODE + \angle EOD = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle EOD \\[3ex] 42.5 + 42.5 + \angle EOD = 180 \\[3ex] 85 + \angle EOD = 180 \\[3ex] \angle EOD = 180 - 85 \\[3ex] \angle EOD = 95^\circ \\[3ex] \implies \\[3ex] \overset{\huge\frown}{DE} = \angle EOD = 85^\circ $
