Difference between revisions of "1954 AHSME Problems/Problem 31"
(Created page with "== Problem== In <math>\triangle ABC</math>, <math>AB=AC</math>, <math>\angle A=40^\circ</math>. Point <math>O</math> is within the triangle with <math>\angle OBC \cong \angle...") |
|||
| (2 intermediate revisions by the same user not shown) | |||
| Line 7: | Line 7: | ||
==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>. | ||
==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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
