In triangle ABC, sinA=0.6, a=10 and b=7. Find sinB
sin A/a = sin B/b law of sines
so
sinB = (b/a) sinA
= .7 sin A = .7*.6 = .42
To find sinB, we can use the sine rule, which states that in any triangle ABC:
a/sinA = b/sinB = c/sinC
Given that sinA = 0.6, a = 10, and b = 7, we can plug these values into the sine rule equation:
10/0.6 = 7/sinB
To isolate sinB, we can cross multiply:
10 * sinB = 7 * 0.6
To solve for sinB, divide both sides by 10:
sinB = (7 * 0.6) / 10
sinB = 0.42