Question

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

Answer 1

`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:

`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"`