Difference between revisions of "1971 AHSME Problems/Problem 5"
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== Solution 1 == | == Solution 1 == | ||
+ | We see that the measure of <math>P</math> equals <math>(\widehat{BD}-\widehat{AC})/2</math>, and that the measure of <math>Q</math> equals <math>\widehat{AC}/2</math>. | ||
+ | Since <math>\widehat{BD} = \widehat{BQ} + \widehat{QD} = 42^{\circ} + 38^{\circ} = 80^{\circ}</math>, the sum of the measures of <math>P</math> and <math>Q</math> is <math>\widehat{BD}/2 = 80^{\circ}/2 = 40^{\circ} \Longrightarrow \boxed{\textbf{(C) }40^{\circ}}</math>. | ||
− | + | == Solution 2 == | |
− | + | Arcs are measured by the angle measures of their corresponding [[central angle|central angles]]. Thus, the [[inscribed angle]] <math>\measuredangle BAQ = \tfrac{42^{\circ}}2 = 21^{\circ}</math>, and, likewise, <math>\measuredangle QCD = \tfrac{38^{\circ}}2 = 19^{\circ}</math>. Thus, by [[supplementary]] angles, <math>\measuredangle PAQ = 180^{\circ} - \measuredangle BAQ = 159^{\circ}</math>, and <math>\measuredangle PCQ = 180^{\circ} - \measuredangle QCD = 161^{\circ}</math>. Because the sum of the [[interior angle]] measures of a quadrilateral add to <math>360^{\circ}</math>, we see that <math>\measuredangle P + \measuredangle Q = 360^{\circ} - \measuredangle PAQ - \measuredangle PCQ = 360^{\circ} - 159^{\circ} - 161^{\circ} = 40^{\circ}</math>. Thus, our answer is <math>\boxed{\textbf{(C) }40^{\circ}}</math>. | |
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 10:03, 1 August 2024
Contents
Problem 5
Points , and
lie on the circle shown and the measures of arcs
and
are
and
respectively. The sum of the measures of angles
and
is
Solution 1
We see that the measure of equals
, and that the measure of
equals
.
Since
, the sum of the measures of
and
is
.
Solution 2
Arcs are measured by the angle measures of their corresponding central angles. Thus, the inscribed angle , and, likewise,
. Thus, by supplementary angles,
, and
. Because the sum of the interior angle measures of a quadrilateral add to
, we see that
. Thus, our answer is
.
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
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