< = angles
determine how many solution exist
<A = 65 , a = 10 , b=20
Solutions ----
< B = __ <C = ___ c=__
<B= __ <C=__ c=__
< = angles
determine how many solution exist
<A = 65 , a = 10 , b=20
Solutions ----
< B = __ <C = ___ c=__
<B= __ <C=__ c=__
I think you may have confused the measure of sides a and b and switched them, so I'll assume that a= 20 and b = 10
Recall the law of sines, which states the following: for ΔABC, with sides a, b, and c
sin(A)/a = sin(B)/b = sin(C)/c
Given the measure of angle A (65°) and side lengths a (20) and b (10), then we can use the law of sines to find the measure of angle B first:
sin(65)/20 = sin(B)/10
10(sin(65)) = 20(sin(B))
10(sin(65))/20 = sin(B)
sin(B) = sin(65)/2
B = sin^{-1}(sin(65)/2)
B = 26.9 ≈ 27
Now that we have the measures of angle A (65°) and angle B (27°), then we can find the measure of angle C from the fact that the sum of the measures of the angles of a triangle is equal to 180°:
65 + 27 + C = 180
C = 180 - 65 - 27
C = 88
Thus, with the measure of angle C (88°), we can again use the law of sines to determine side c:
sin(65)/20 = sin(88)/c
c(sin(65)) = 20(sin(88))
c = 20(sin(88))/(sin(65))
c = 22.05 ≈ 22
Thus, side c is equal to 22.
In triangle ABC, sin(B)/20 = sin(65)/10
Solve for sin(B),
sin(B) = 1.813, which is impossible. So, there is no solution for the problem.
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Attn: I assumed the question was about a triangle.