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

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

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Answer 1 · 2018-05-06T09:52:46

$C = 90^{\circ}$ $C=90^circ$

$B = 90^{\circ} - 46^{\circ} = 44^{\circ} .$ $B=90^circ-46^circ=44^circ.$

$b = \frac{a}{tan A} \approx 30.902$ $b = a/tan A approx 30.902$

$c = \frac{a}{sin A} \approx 44.485$ $c = a/sin A approx 44.485$

Explanation:

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

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

$tan A = \frac{a}{b}$ $tan A = a/b$

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

$sin A = \frac{a}{c}$ $sin A = a/c$

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

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

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