SOLUTION: Help me figure this out,
In a triangle, two sides that measure 6 cm and 10 cm form an angle that measures 80 degrees. Find, to the nearest degree, the measure of the smallest ang
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-> SOLUTION: Help me figure this out,
In a triangle, two sides that measure 6 cm and 10 cm form an angle that measures 80 degrees. Find, to the nearest degree, the measure of the smallest ang
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Question 790000: Help me figure this out,
In a triangle, two sides that measure 6 cm and 10 cm form an angle that measures 80 degrees. Find, to the nearest degree, the measure of the smallest angle in the triangle. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! use the law of cosines to find the length of the third side
c^2 = 10^2 + 6^2 - 2*10*6(cos(80))
c^2 = 136 -120*.17
c^2 = 136 -20.84
c = 10.7
now use the law of sines
a/sin(A) = b/sin(B) = c / sin(C)
10.7 / sin(80) = 10 / sin(B) = 6 / sin(C)
sin(B) = 10 / 10.7 = .934
sin(69) = .934
now
80 + 69 = 149 and
180 - 149 = 31
therefore the smallest angle is 31 degrees
