# triangle, cosines to solve

Given a triangle ABC with the information

a=59 inches    b=70 inches    c=111inches

Use the law of Cosines to solve the triangle

asked Feb 23, 2013 in CALCULUS

Given that a = 59 inches, b = 70 inches and c = 111 inches.

The law of cosines are

a2 = b2 + c2 - 2bc[cos(A)]

b2 = a2 + c2 - 2ac[cos(B)]

c2 = a2 + b2 - 2ab[cos(C)]

Take a2 = b2 + c2 - 2bc[cos(A)]

Substitute a = 59, b = 70 and c = 111 in the cosine law.

592 = 702 + 1112 - 2(70)(111)[cos(A)]

3481 = 4900 + 12321 - 15540[cos(A)]

3481 = 17221 - 15540[cos(A)]

Subtract 17221 from each side.

-13740 = - 15540[cos(A)]

Divide each side by -15540.

[cos(A)] = 0.8841

Trigonometric table in cos(27°.9')

cos(A) = cos(27°.9') then A = 27°.9'

Take b2 = a2 + c2 - 2ac[cos(B)]

Substitute a = 59, b = 70 and c = 111 in the cosine law.

702 = 592 + 1112 - 2(59)(111)[cos(B)]

4900 = 3481 + 12321 - 13098[cos(B)]

4900 = 15802 - 13098[cos(B)]

Subtract 15802 from each side.

-10902 = - 13098[cos(B)]

Divide each side by -13098.

[cos(B)] = 0.8323

Trigonometric table in cos(33°.6')

cos(B) = cos(33°.6') then B = 33°.6'

Take c2 = a2 + b2 - 2ab[cos(C)]

Substitute a = 59, b = 70 and c = 111 in the cosine law.

1112 = 592 + 702 - 2(59)(70)[cos(C)]

12321 = 3481 + 4900 - 8260[cos(C)]

12321 = 8381 - 8260[cos(C)]

Subtract 8381 from each side.

3490= - 8260[cos(C)]

Divide each side by -8260.

[cos(C)] = - 0.4225

Trigonometric table in cos(115°)

cos(C) = cos(115°) then C = 115°

Therefore A = 27°.9', B = 33°.6' and C = 115°.