Answer:
A. The length of the third side is approximately 5 feet.
B. A = [tex]88^{o}[/tex], B = [tex]60^{o}[/tex] and C = [tex]32^{o}[/tex].
Step-by-step explanation:
Let the triangle be ABC. Given two sides and an included angle, let us apply the cosine rule to determine the third length.
A. Let side a = 6 feet and c = 3 feet, thus;
[tex]b^{2}[/tex] = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] - 2ac Cos B
= [tex]6^{2}[/tex] + [tex]3^{2}[/tex] - 2(6 x 3) Cos [tex]60^{o}[/tex]
= 36 + 9 - 36 x 0.5
= 45 - 18
[tex]b^{2}[/tex] = 27
b = [tex]\sqrt{27}[/tex]
= 5.1962
b = 5.2 feet
b ≅ 5 feet
The length of the third side is approximately 5 feet.
B. Given that B = [tex]60^{o}[/tex], then let us apply the Sine rule to determined the measure of A.
[tex]\frac{a}{SinA}[/tex] = [tex]\frac{b}{SinB}[/tex]
So that,
[tex]\frac{6}{Sin A}[/tex] = [tex]\frac{5.2}{Sin60^{o} }[/tex]
Sin A = [tex]\frac{0.886*6}{5.2}[/tex]
Sin A = 0.99923
A = [tex]Sin^{-1}[/tex] 0.99923
= 87.75
A ≅ [tex]88^{o}[/tex]
Since the sum of angle in a triangle = [tex]180^{o}[/tex].
Then,
A + B + C = [tex]180^{o}[/tex]
[tex]88^{o}[/tex] + [tex]60^{o}[/tex] + C = [tex]180^{o}[/tex]
[tex]148^{o}[/tex] + C = [tex]180^{o}[/tex]
C = [tex]180^{o}[/tex] - [tex]148^{o}[/tex]
= [tex]32^{o}[/tex]
C = [tex]32^{o}[/tex]
Thus,
A = [tex]88^{o}[/tex], B = [tex]60^{o}[/tex] and C = [tex]32^{o}[/tex].
