a ambiguous case in Trigonometry?

Question

solve the triangle with a=7, b=10, and c=6, and find its area. Round all answers to the nearest tenth.
I slove all the angles by the law of cosine. (though I can do it with the law of sin)
then I got A=43.5
B=79.8
C=36.2
I have checked the value of B and C is correct, what's wrong with A of my answer?
please explain by using the law of cosine

Answer 1

Actually, your value for A is perfectly correct. It's your value for ∠B that's wrong. It should be 100.3°. Consider:

cos B = (a²+c²-b²)/(2ac) = (49+36-100)/84 = -15/84 = -5/28
B = arccos (-5/28) ≈ 100.3°

My guess is that you dropped the negative sign when you plugged the value for cos B into your calculator -- that would cause you to get 79.8.

Answer 2

I dont think A is off, i think B is be off.
a = 7, b = 10, c = 6

b^2 = a^2 + c^2 - 2ac Cos B
(10)^2 = (7)^2 + (6)^2 - 2(7)(6) Cos B
100 = 49 + 36 - 84 Cos B
100 = 85 - 84 Cos B
15 = -84 Cos B
Cos B = 15 / -84
B = Cos^-1(15 / - 84)
B = 100.29 degrees.

If you had done the last step without the negative it would be:
B = Cos^-1 (15 / 84)
B = 79.71 degrees.

Just to check A;
a^2 = b^2 + c^2 - 2bc Cos A
7^2 = 10^2 + 6^2 - 2(10)(6) Cos A
49 = 100 + 36 - 120 Cos A
49 = 136 - 120 Cos A
-87 = - 120 Cos A
Cos A = ( - 87 / - 120)
A = Cos^-1 (87/120)
A = 43.53 degrees.

And you can just find C by subtracting A and B from 180. (180 - 43.53 - 100.29 = 36.18 degrees)

So your values for A and C were correct, it was just your value for B that was off.

Because you ask to show with law of cosines, i will show angle C with law of cosines.

c^2 = a^2 + b^2 - 2ab Cos C
6^2 = 7^2 + 10^2 - 2(7)(10) Cos C
36 = 49 + 100 - 140 Cos C
36 = 149 - 140 Cos C
- 113 = - 140 Cos C
Cos C = -113 / -140
C = Cos^-1(113 / 140)
C = 36.18 degrees.

It looks like all you did was forget the negative with B.

Answer 3

once you're utilising the regulation of sines to remedy a triangle, now and back there are 2 achievable recommendations. enable A = 35 ranges, a = 10, and b = 15. sin A / a = sin B / b sin 35 / 10 = sin B / 15 sin B = 15 sin 35 / 10 sin B = 0.8604 B could nicely be the two fifty 9.4 ranges or one hundred twenty.6 ranges, so there are 2 achievable recommendations for the triangle. One is acute, and one is obtuse.

Answer 4

A+B+C must be 180 degrees.
Probably that is how you decided your answer was off.

Answer 5

try this tool :D http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html