How do you use the law of sines to solve the triangle given A = 75.1°, B = 37.9°, b = 30?

Question

How do you use the law of sines to solve the triangle given A = 75.1°, B = 37.9°, b = 30?

1 Answer

Impact of this question

1982 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-06T15:52:59

$C = 180^{\circ} - A - B$ $C = 180^circ - A - B$

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

$c = \frac{b sin C}{sin B}$ $c = {b sin C}/sin B$

Explanation:

$C = 180^{\circ} - A - B = 180^{\circ} - {75.1}^{\circ} - {37.9}^{\circ} = 67^{\circ}$ $C = 180^circ - A - B = 180^circ - 75.1^circ - 37.9^circ = 67^circ$

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sin A = b/sin B = c/sin C$

$a = b \frac{sin A}{sin B} = 30 \frac{{sin 75.1}^{\circ}}{{sin 37.9}^{\circ}} \approx 47.1952$ $a = b sin A/sin B = 30 { sin 75.1 ^circ }/{ sin 37.9^circ } approx 47.1952$

$c = b \frac{sin C}{sin B} = 30 \frac{{sin 67}^{\circ}}{{sin 37.9}^{\circ}} \approx 44.9549$ $c = b sin C/sin B = 30 { sin 67^circ }/{ sin 37.9^circ } approx 44.9549$

Science

Math

Humanities

... and beyond

How do you use the law of sines to solve the triangle given A = 75.1°, B = 37.9°, b = 30?

1 Answer

Explanation:

Related questions

Impact of this question