< = 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=__

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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.