+188 A triangle has side lengths 10, 15, and 7. Is the triangle acute, obtuse, or right?
+129899 This might be a little above your level, CarpeDiem....but.......we can find the angle opposite the longest side and determine what the correct answer is........this is done with the Law of Cosines.......we have..... 10,15,7
15^2 = 10^2 + 7^2 - 2(10)(7)cosA where A is the angle opposite the longest side.......rearrange as:
cosA = [15^2 - 10^2 - 7^2] / [-2(10)(7)]
And using the cosine inverse [ the arccos], we have
arccos ( [15^2 - 10^2 - 7^2] / [-2(10)(7)] ) = A = about 122.9°
So.....this triangle is obtuse
BTW.....here's a handy-dandy resource that will let you calculate triangles yourself...
http://www.mathwarehouse.com/triangle-calculator/online.php
+129899 This might be a little above your level, CarpeDiem....but.......we can find the angle opposite the longest side and determine what the correct answer is........this is done with the Law of Cosines.......we have..... 10,15,7
15^2 = 10^2 + 7^2 - 2(10)(7)cosA where A is the angle opposite the longest side.......rearrange as:
cosA = [15^2 - 10^2 - 7^2] / [-2(10)(7)]
And using the cosine inverse [ the arccos], we have
arccos ( [15^2 - 10^2 - 7^2] / [-2(10)(7)] ) = A = about 122.9°
So.....this triangle is obtuse
BTW.....here's a handy-dandy resource that will let you calculate triangles yourself...
http://www.mathwarehouse.com/triangle-calculator/online.php
