How do you solve the $triangle ABC$ given $A=50^circ, a=2.5, C=67^circ$ ?

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Answer 1 · 2017-08-05T19:55:56

See below.

Since all the angles of a triangle sum to $180^@$ , we can say

$m /_ A + m /_ B + m /_ C = 180^@$

Substitute in the given information and solve for $m/_B$ :

$50^@ + m /_ B + 67^@ = 180^@$

$m /_ B + 117^@ = 180^@$

$m /_ B = 180^@ - 117^@$

$m /_ B = 63^@$

We now have the following triangle:

![http://mathworld.wolfram.com/Triangle.html (edited by fleur)]( useruploads.socratic.org )

The Law of Sines states that $sin A/a = sin B / b = sin C / c$ .

Thus, we can say

$sin 50^@/2.5 = sin 63^@/b$

Cross-multiply and solve for $b$ :

$b * sin 50^@ = 2.5 * sin 63^@$

$b = (2.5 * sin 63^@)/sin 50^@$

$b=2.9$

Do the same to find $c$ :

$sin 50^@/2.5 = sin 67^@/c$

$c * sin 50^@ = 2.5 * sin 67^@$

$c = (2.5 * sin 67^@)/sin 50^@$

$c=3.0$

$m/_A=50^@$
$m/_B=63^@$
$m/_C=67^@$

$a=2.5$
$b=2.9$
$c=3.0$

How do you solve the triangle ABCtriangle ABC given A=50^circ, a=2.5, C=67^circA=50^circ, a=2.5, C=67^circ?