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.