How do you solve the triangle given #C=103^circ, a=3/8, b=3/4#?

1 Answer
Mar 25, 2018

Answer:

#color(purple)(c = 0,9109, hat A = 23.65^@, hat B = 53.35^@#

#color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = 0.137 "sq units"#

Explanation:

http://www.dummies.com/education/math/trigonometry/laws-of-sines-and-cosines/

#hat C = 103^@, a = 3/8, b = 3/4, " To solve the triangle"#

Applying Law of cosines,

#c = sqrt(a^2 + b^2 - 2 a b cos C)#

#c = sqrt((3/8)^2 + (3/4)^2 - (2 * (3/8) * (3/4) * cos 103))#

#c = 0.9109#

Applying Law of sines,

# sin A / (3/8) = sin B / (3/4) = sin 103 / 0.9109#

#sin A = ((3/8) * sin 103) / 0.9109 = 0.4011#

#hat A = sin ^-1 0.4011 = 23.65^@#

Similarly, #sin B = ((3/4) * sin 103) / 0.9109 = 0.8023#

#hat A = sin ^-1 0.8023 = 53.35^@#

#color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = (1/2) * (3/8) * (3/4) * sin 103 = 0.137 "sq units"#