Question

Solve the triangle.

B = 36°, a = 41, c = 20

Answer 1

Answer:

See below

Step-by-step explanation:

When you have 2 sides and the angle between them you use the cosine theorem or law of cosines:

$b^{2}=a^{2}+c^{2}-2acCos(B)$

$b^{2}=41^{2}+20^{2}-2(41)(20)cos36 \\b^{2}=1681 + 400-1327 \\b^{2}=754 \\b = 27.5$

After you have a side and the opposed angle (side b and angle B), you use the law of sines:

$\frac{a}{sinA} =\frac{b}{sinB}=\frac{c}{sinC}$

I will calculate angle A first:

$[tex]\frac{27.5}{sin36} =\frac{41}{sinA} \\sinA =\frac{41}{27.5} sin36 \\sin A = 0.876 \\A = 61.2\°$[/tex]

Same for angle C:

$\frac{27.5}{sin36} =\frac{20}{sinC} \\sinC =\frac{20}{27.5} sin36 \\sin C = 0.427 \\C = 25.3\°$

The sum of the angles is 36° + 61.2° + 25.3° = 122.5°; The sum must be 180° so this isnt a triangle.

I re did the problem using B as 63° instead of 36° in case you wrote it wrong and I got a satisfactory answer.

using B = 36:

b = 36.6

angle A = 86.4°

angle C = 29.1°

A+B+C = 86.4° + 29.1° + 63° = 178.5° ~ 180°