Measure of the smallest angle in the triangle

priscilla98
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Homework Statement



In a triangle, two sides that measure 6 centimeters and 10 centimeters form an angle that measures 80 degrees. Find to the nearest degree, the measure of the smallest angle in the triangle.


Homework Equations



A2 = B2 + C2 – 2bc Cos A

The Attempt at a Solution



I tried using this formula to get the answer but it didn't work out. This is one of the questions from the algebra 2 trigonometry regents. I know the answer which is 33 degrees but I don't know on how to get to that answer.
 
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I think relevant equation would actually be

\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

along with the equation you've mentioned
 


Okay, then a = 6, b = 10 and C = 80 degrees, right? Then idk how to use the revelant equations you implied.
 


priscilla98 said:
Okay, then a = 6, b = 10 and C = 80 degrees, right? Then idk how to use the revelant equations you implied.

i just edited my reply. use your equation first and find c. Then use my equation to get A and B
 


Can you show us your work? Because I was able to get the answer (approx 33.409°). Make sure you are in degree mode when doing your calculations (or convert the 80° to radians).

Perhaps using the Law of Sines will work, but the OP may be required to use Law of Cosines for this problem.69

EDIT: Too slow ^.^
 


Thanks a lot praharmitra for this. I really appreciate it.
 


Actually, if you're not sure about the sine law, you can solve it using sin, cos and tan in right-angled triangles (draw the triangle, call its height h and relate it to the sine of 80 degrees, then divide the side of length 10 into two pieces of length L and 10 - L and relate L to the cosine of 80 degrees).
 


first you need to find side c using cosine law c2=a2 +b2=2abcosC
then use sine law to find angle A, cause the higher angle, the longer the side, since side A is the shortest, which means angle A is the smallest.
 

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