How do you solve triangle ABC, given B=41°, c=7, a=10?

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Answer 1 · 2015-05-30T03:07:49

Use the trig identities: # b^2 = a^2 + c^2 - 2ac.cos B #

and #sin A/a = sin B/b.#

#b^2 = 100 + 49 - 140(0.75) = 149 - 105 = 44 -> b = 6.63#

#sin A/ 10 = sin (41)/6.63 # ->

#sin A = (10(0.66))/6.63 = 1 -> A = 90# deg

C = 90 - B = 90 - 41 = 49 deg

Answer 2 · 2015-05-30T03:37:31

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Hope this helps!