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Difference between revisions of "1956 AHSME Problems/Problem 49"

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Solution 1
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==Problem==
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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 1==
  
 
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>.

Revision as of 18:55, 18 July 2021

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 1

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 \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]

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