Difference between revisions of "1954 AHSME Problems/Problem 31"
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==Solution== | ==Solution== | ||
| − | Since <math>\triangle ABC</math> is an isosceles triangle, <math>\angle ABC = \angle ACB = 70^{\circ}</math>. Let <math>\angle OBC = \angle OCA = x</math>. Since <math>\angle ACB = 70</math>, <math>\angle OCB = 70 - x</math>. The angle of <math>\triangle OBC</math> add up to <math>180</math>, so <math>\angle BOC = 180 - (x + 70 - x) = \boxed{\textbf{(A) } 110^{\circ}</math>. | + | Since <math>\triangle ABC</math> is an isosceles triangle, <math>\angle ABC = \angle ACB = 70^{\circ}</math>. Let <math>\angle OBC = \angle OCA = x</math>. Since <math>\angle ACB = 70</math>, <math>\angle OCB = 70 - x</math>. The angle of <math>\triangle OBC</math> add up to <math>180</math>, so <math>\angle BOC = 180 - (x + 70 - x) = \boxed{\textbf{(A) } 110^{\circ}}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 00:51, 28 February 2020
Problem
In 
, 
, 
. Point 
is within the triangle with 
.
The number of degrees in 
is:
Solution
Since is an isosceles triangle, 
. Let 
. Since 
, 
. The angle of 
add up to 
, so 
.
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
