Consider a triangle ABC like the one below Suppose that c 5

Consider a triangle ABC like the one below. Suppose that c = 50, b = 57, and C = 33degree. (The figure is not drawn to scale.) Solve the triangle. carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. If no such triangle exists, enter \"No solution.\" If there is more than one solution, use the \"or\" button.

Solution

we know two side lengths b and c , an angle then we can apply cosine forula

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

given c=50 , b=57 C=33 degree

(50)^2 = a^2 + (57)^2 - 2a(57) cos(33)

2500 = a^2 +3249 -114a x0.84

a^2 - 95.76a + 3249-2500=0

a^2 -95.76a +749 =0   

a^2 -95.76a+749=0

a = 87.167 or 8.6

but we will take a= 87.167

now we know all the side lengths

then by applying the sine law we can get A and B angles also

a/sinA =b/sinB =c/sinC

87.2 / sinA = 50/sin(33)

(87.2/50) sin(33) = sin(A)

sin(A) = 1.744x0.545

sinA= 0.95

A= arc sin(0.95)

A=71.89 degree = 72 degree

we know A and C then we can find B by the sum of all angles in a triangle is 180

A+B+C=180

72+B+33 =180

B = 180-105

B=75 degree