How to use the sine law and cosine law to solve a triangle problem?

Question:

In the figure shown, ABC is a triangle. BA = BD = a (a is a number.) AB = Sqrt (2) × AD, and 3 × AB = Sqrt (2) × AC. Find the degree measure of the angle C.

Solution:

Because we know the three sides of the triangle ABD and we also know the side AC which opposite the angle B in triangle ABC, we need to find the angle B first.

cos B = BA^{2} + BD^{2} - AD^{2} ÷ [2 × BA × BD]

we are given:

AB = BD = a

AB = Sqrt (2) AD, so, AD = AB/Sqrt (2) = a/Sqrt (2)

3 AB = Sqrt (2) AC, so, AC = 3 AB/Sqrt (2) = 3a/Sqrt (2)

cos B = [a^{2} + a^{2} - (a/(Sqrt (2)))^{2}] / [2 × a × a]

= [a^{2} + a^{2} - a^{2}/2] / 2a^{2}

= a^{2}(1 + 1 - 1/2)/2a^{2}

= (2 - 1/2)/2

= 3/4

find the value of sin B

sin^{2}B = 1 - cos^{2}B

= 1 - (3/4)^{2}

= 1 - 9/16

= 7/16

sin B = Sqrt (7)/4

In triangle ABC, we know the angle B and its opposite side AC, and we also know the side AB which opposite the angle C. Use the sine law to find the angle C.

In triangle ABC,

sin C/AB = sin B/AC

sin C = (AB/AC) sin B

= a/[3a/Sqrt (2)] × (Sqrt (7))/4

= Sqrt (2)/3 × Sqrt (7)/4

= Sqrt (14)/12

= 0.3118

angle C = arc sin 0.3118 = 18.17^{o}

Therefore, the angle C is 18.17 degrees, please watch the video for more details.