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A triangle has side lengths 10, 15, and 7. Is the triangle acute, obtuse, or right?

 Apr 6, 2016

Best Answer 

 #1
avatar+129849 
+5

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

 

 

 

 

 

cool cool cool

 Apr 6, 2016
 #1
avatar+129849 
+5
Best Answer

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

 

 

 

 

 

cool cool cool

CPhill Apr 6, 2016
 #2
avatar+188 
0

Thank you, thank you, thank you!

Carpe♥Diem  Apr 6, 2016

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