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### 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 = BA2 + BD2 - AD2 ÷ [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 = [a2 + a2 - (a/(Sqrt (2)))2] / [2 × a × a]
= [a2 + a2 - a2/2] / 2a2
= a2(1 + 1 - 1/2)/2a2
= (2 - 1/2)/2
= 3/4

find the value of sin B

sin2B = 1 - cos2B
= 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 = 18.17o

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