Consider a triangle ABC like the one below Suppose that c 4

Consider a triangle ABC like the one below. Suppose that c = 43, b = 69, and C = 27 degree. (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

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

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

-a^2 = 69^2 - 43^2 - 2 a ( 69 ) cos( 27)

-a^2 = 2912 - 2 a ( 69 ) 0.891

-a^2 + 122.958 a - 2912= 0

a^2 - 122.958 a + 2912= 0

a = 90.935 , 32.02250

therefore a = 20.3231

a^2 = b^2 + c^2 - 2 b c cos ( A )

2 b c cos ( A ) = b^2 + c^2 - a^2

cos ( A ) =( 69^2 + 43^2 - (90.9)^2 )/2*43*69

cos( A ) = -0.27853

A = 105.664

b^2 = a^2 + c^2 - 2ac cos ( B )

==> 2ac cos ( B ) = a^2 + c^2 - b^2

==> cos ( B ) = ((90.9)^2 + 43^2 - 69^2 )/2*90.9*43

==> cos ( B ) = 0.68447

===> B = 46.80