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Difference between revisions of "1956 AHSME Problems/Problem 49"
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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 180^\circ - \angle BAO - \angle ABO =
+
<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}=  
Line 18: Line 18:
  
 
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>
 +
 
== See Also ==
 
== See Also ==
  

Latest revision as of 00:32, 3 January 2024

Problem

Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40^{\circ}$; then $\angle AOB$ equals:

$\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}$


Solution

First, from triangle $ABO$, $\angle AOB = 180^\circ - \angle BAO - \angle ABO$. Note that $AO$ bisects $\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\angle BAO = \angle BAT/2$. Similarly, $\angle ABO = \angle ABR/2$.

Also, $\angle BAT = 180^\circ - \angle BAP$, and $\angle ABR = 180^\circ - \angle ABP$. Hence,

$\angle AOB = 180^\circ - \angle BAO - \angle ABO = 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} = 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}= \frac{\angle BAP + \angle ABP}{2}.$

Finally, from triangle $ABP$, $\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ$, so \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]

See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 48 		Followed by
Problem 50
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All AHSME Problems and Solutions

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