How would you find the sides of a triangle given angle A=50, angle B=30, and side a=1?

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Answer 1 · 2018-05-02T05:52:15

$a = 1$ (given)
$b = 0.65$
$c = 1.29$

To solve this, we use the Law of Sines

$b/sinB = a/sinA$

$b/(sin30^@) = 1/sin50^@$

$b = sin30^@/sin50^@$

$b ~~ 0.65$

$-------------------$

We can find angle C by doing $180^@ - 50^@ - 30^@ = 100^@$

Now let's find side c:
$c/sinC = a/sinA$

$c/(sin100^@) = 1/sin50^@$

$c = sin100^@/sin50^@$

$c ~~ 1.29$

Therefore, the lengths of the sides of the triangle are:
$a = 1$ (given)
$b = 0.65$
$c = 1.29$

( $b$ and $c$ are rounded to the nearest hundredth)

Hope this helps!