1 Given oblique ABC with A110 B 40 and a 30 solve the tria

1) Given oblique ABC with A-110°, B = 40°, and a = 30, solve the triangle. A)C = 30°, c = 20.521, b = 15.963 B) C = 30°, c = 15.963, b = 20.521 C) C = 30°, c = 20.294, b = 15.864 D)C = 30°, c = 15.864, b = 20.294 E) C = 30°, c = 75.963, b = 67.193 2) Determine the equation of the line that passes throu and is perpendicular to y =2x + 6. gh (-1, 2) By-2--(x+1) E) y-2=--(x +1) V3, 3) Determine arcsin(-2). E)-

Solution

1) Option B) is correct.

A = 110,   B = 40,    a = 30

C = 180 – 110 – 40 = 30 degree

Now

sinC/c = sinA/a =======> sin30/c = sin110/30 ========> c = 15.963

and

sinB/b = sinA/a =======> sin40/b = sin110/30 ========> c = 20.521

3) Option E) is correct.

arcsin(-sqrt(3)/2) = -arcsin(sqrt(3)/2) = -pi/3