Solve the triangles with A 32 degree a 10 and b 8Solution

Solve the triangle(s) with A = 32 degree, a = 10, and b = 8.

Solution

A = 32 deg ; a= 10 and b =8

use sine law: a/sinA = b/sinB

10/sin32 = 8/sinB

sinB = 8*sin32/10 = 0.4239

B1 = 25.08 deg ; B2 = 180 - 25.08 = 154.9 deg

C1 = 180 - A - B1 = 122.92 deg ; C2 = 180 - A - B2 = -6.9 deg Not possible

Only 1 triangle is possible A= 32 ; B1 = 25.08 deg; C1 = 122.92 deg

c/sinC = a/sinA

c = sin122.92*10/sin32 = 15.97

 Solve the triangle(s) with A = 32 degree, a = 10, and b = 8.SolutionA = 32 deg ; a= 10 and b =8 use sine law: a/sinA = b/sinB 10/sin32 = 8/sinB sinB = 8*sin32/

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