How do you solve the right triangle ABC if a=32m, A=46?

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Answer 1 · 2018-05-06T09:52:46

$C=90^circ$

$B=90^circ-46^circ=44^circ.$

$b = a/tan A approx 30.902$

$c = a/sin A approx 44.485$

$a=32$ m, $A=46^circ.$ Let's say $C=90^circ$ so $B=90^circ-46^circ=44^circ.$

This is almost an isosceles triangle but it's certainly a right triangle. We have

$tan A = a/b$

$b = a/tan A = 32/{tan 46^circ} approx 30.902$

$sin A = a/c$

$c = a/sin A = 32/{sin 46^circ} approx 44.485$