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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 = (square root of 2) × AD, and 3 × AB = (square root of 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 = (square root of 2) AD, so, AD = AB/(square root of 2) = a/(square root of 2)
- 3 AB = (square root of 2) AC, so, AC = 3 AB/(square root of 2) = 3a/(square root of 2)

- cos B = [a
^{2}+ a^{2}- (a/(square root of 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 = (square root of 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/(square root of 2)] × (square root of 7)/4
- = (square root of 2)/3 × (square root of 7)/4
- = (square root of 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.