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

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

To solve this, we use the Law of Sines

$\frac{b}{sin B} = \frac{a}{sin A}$ $b/sinB = a/sinA$

$\frac{b}{{sin 30}^{\circ}} = \frac{1}{{sin 50}^{\circ}}$ $b/(sin30^@) = 1/sin50^@$

$b = \frac{{sin 30}^{\circ}}{{sin 50}^{\circ}}$ $b = sin30^@/sin50^@$

$b \approx 0.65$ $b ~~ 0.65$

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

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

Now let's find side c:
$\frac{c}{sin C} = \frac{a}{sin A}$ $c/sinC = a/sinA$

$\frac{c}{{sin 100}^{\circ}} = \frac{1}{{sin 50}^{\circ}}$ $c/(sin100^@) = 1/sin50^@$

$c = \frac{{sin 100}^{\circ}}{{sin 50}^{\circ}}$ $c = sin100^@/sin50^@$

$c \approx 1.29$ $c ~~ 1.29$

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

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

Hope this helps!