Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 10"
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== Solution == | == Solution == | ||
− | To find BD, we need to first find the altitude. If we have the angles of the triangle, we can use law of sines to find the length of the altitude. We know that the total degrees in a triangle is 180, so we can use law of sines to set up a proportion. The sum of the side lengths is 156. Diving 180 by 156 gives us each 15/13 for each side unit. | + | To find BD, we need to first find the altitude. If we have the angles of the triangle, we can use law of sines to find the length of the altitude. We know that the total degrees in a triangle is 180, so we can use law of sines to set up a proportion. The sum of the side lengths is 156. Diving 180 by 156 gives us each 15/13 for each side unit. Multiplying by 53, we get angle B is 795/13 degrees. |
== See also == | == See also == |
Revision as of 22:42, 23 December 2014
Problem
The scalene triangle has side lengths is perpendicular to
(a) Determine the length of
(b) Determine the area of triangle
Solution
To find BD, we need to first find the altitude. If we have the angles of the triangle, we can use law of sines to find the length of the altitude. We know that the total degrees in a triangle is 180, so we can use law of sines to set up a proportion. The sum of the side lengths is 156. Diving 180 by 156 gives us each 15/13 for each side unit. Multiplying by 53, we get angle B is 795/13 degrees.
See also
1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |