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