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: $b^{2} = a^{2} + c^{2} - 2 a c . cos B$ $b^2 = a^2 + c^2 - 2ac.cos B$

and $\frac{sin A}{a} = \frac{sin B}{b} .$ $sin A/a = sin B/b.$

$b^{2} = 100 + 49 - 140 (0.75) = 149 - 105 = 44 \to b = 6.63$ $b^2 = 100 + 49 - 140(0.75) = 149 - 105 = 44 -> b = 6.63$

$\frac{sin A}{10} = \frac{sin (41)}{6.63}$ $sin A/ 10 = sin (41)/6.63$ ->

$sin A = \frac{10 (0.66)}{6.63} = 1 \to A = 90$ $sin A = (10(0.66))/6.63 = 1 -> A = 90$ deg

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

Answer 2 · 2015-05-30T03:37:31

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Hope this helps!