Question

Identify all solutions for a triangle with A - 38°, b= 10, and a=8. Round to the nearest tenth.

Answer 1

Answer:

A=38 degrees, B=50 degrees, C=92 degrees

a=8, b=10, c=13

Step-by-step explanation:

Given a triangle where: A=38°, b= 10, and a=8.

Using Law of Sines

$\dfrac{a}{sin A} =\dfrac{b}{sin B} \\\dfrac{8}{sin 38} =\dfrac{10}{sin B} \\ Cross multiply\\8 X sin B=10 X Sin 38\\Sin B=(10 X sin 38)\div 8\\B=arcsin[(10 X sin 38)\div 8]\\B\approx50^\circ$

$\angle A+\angle B+\angle C=180^\circ( Sum of angles in a triangle)\\38+50+\angle C=180^\circ\\\angle C=180-88\\\angle C=92^\circ$

$\dfrac{a}{sin A} =\dfrac{c}{sin C} \\\dfrac{8}{sin 38} =\dfrac{c}{sin 92} \\ Cross multiply\\8 X sin 92=c X Sin 38\\c=(8 X sin 92)\div sin38\\c\approx13$

Answer 2

Answer:

B= 50.35°

C=91.65°

c= 12.77

Step-by-step explanation:

Given:

A = 38°

b= 10 and a=8.

Required:

angles B and C, and sides c.

By  using the rule for law of sines

$sin B=b\frac{sinA}{a} = \frac{(10)(0.62)}{8}$ => 0.77

B= $sin^-^1$(0.77) => 50.35°

For angle C:

angle C= 180 - A - B => 180 - 38 - 50.35

            =91.65°

For side c:

c=$a(\frac{sinC}{sinA} )$ => 8($\frac{0.99}{0.62}$)

c= 12.77