Difference between revisions of "1956 AHSME Problems/Problem 49"
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− | Solution | + | ==Problem== |
+ | Triangle <math>PAB</math> is formed by three tangents to circle <math>O</math> and <math>\angle APB = 40^{\circ}</math>; then <math>\angle AOB</math> equals: | ||
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+ | <math>\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}</math> | ||
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+ | ==Solution== | ||
First, from triangle <math>ABO</math>, <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO</math>. Note that <math>AO</math> bisects <math>\angle BAT</math> (to see this, draw radii from <math>O</math> to <math>AB</math> and <math>AT,</math> creating two congruent right triangles), so <math>\angle BAO = \angle BAT/2</math>. Similarly, <math>\angle ABO = \angle ABR/2</math>. | First, from triangle <math>ABO</math>, <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO</math>. Note that <math>AO</math> bisects <math>\angle BAT</math> (to see this, draw radii from <math>O</math> to <math>AB</math> and <math>AT,</math> creating two congruent right triangles), so <math>\angle BAO = \angle BAT/2</math>. Similarly, <math>\angle ABO = \angle ABR/2</math>. | ||
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Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, | Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence, | ||
− | <math>\angle AOB | + | <math>\angle AOB = 180^\circ - \angle BAO - \angle ABO = |
− | 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} | + | 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} = |
− | 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} | + | 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}= |
\frac{\angle BAP + \angle ABP}{2}.</math> | \frac{\angle BAP + \angle ABP}{2}.</math> | ||
Finally, from triangle <math>ABP</math>, <math>\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ</math>, so <cmath>\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.</cmath> | Finally, from triangle <math>ABP</math>, <math>\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ</math>, so <cmath>\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.</cmath> | ||
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+ | == See Also == | ||
+ | |||
+ | {{AHSME 50p box|year=1956|num-b=48|num-a=50}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:32, 3 January 2024
Problem
Triangle is formed by three tangents to circle and ; then equals:
Solution
First, from triangle , . Note that bisects (to see this, draw radii from to and creating two congruent right triangles), so . Similarly, .
Also, , and . Hence,
Finally, from triangle , , so
See Also
1956 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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