How do you solve the right triangle $A B C$ $ABC$ if $a = 20$ $a=20$ m, $B = 21^{\circ}$ $B=21^@$ ?

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How do you solve the right triangle $A B C$ $ABC$ if $a = 20$ $a=20$ m, $B = 21^{\circ}$ $B=21^@$ ?

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Answer 1 · 2015-04-26T16:32:11

$A B C$ $ABC$ if $a = 20$ $a=20$ m, $B = 21^{\circ}$ $B=21^@$

I assume you are using standard labeling, so $C = 90^{\circ}$ $C=90^@$ and $A = 69^{\circ}$ $A=69^@$ and $a$ $a$ is the side opposite $A$ $A$ and so on.

Use ${cos 21}^{\circ} = \frac{20}{c}$ $cos21^@ = 20/c$ to find $c$ $c$ .

and ${cot 21}^{\circ} = \frac{20}{a}$ $cot21^@ = 20/a$ to find $a$ $a$ .

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How do you solve the right triangle $A B C$ $ABC$ if $a = 20$ $a=20$ m, $B = 21^{\circ}$ $B=21^@$ ?

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How do you solve the right triangle ABCABC if a=20a=20m, B=21∘B=21^@?

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How do you solve the right triangle $A B C$ $ABC$ if $a = 20$ $a=20$ m, $B = 21^{\circ}$ $B=21^@$ ?