Difference between revisions of "1971 AHSME Problems/Problem 5"
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− | 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}</math>. | + | 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}, the sum of the measures of < | + | 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 \textbf{(C) }</math>. |
Latest revision as of 22:56, 23 July 2020
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
We see that the measure of equals , and that the measure of equals . Since , the sum of the measures of and is .