In Triangle ABC, AB = 10. AC = 14, and angle A = 51°. Find the length of BC to the nearest hundredth
2 answers:
Answer:
BC = 10.94
Step-by-step explanation:
its a las of cosines
a² = b² + c² - 2bc cos A
AB = 10
AC = 14
angle A = 51°
BC² = 10² + 14² - (2 * 10 * 14 cos 51°)
BC = sqrt 119.8
BC = 10.94
Answer:
BC ≈ 10.94
Step-by-step explanation:
Base on the triangle we were given 2 sides and an angle. We were also asked to find the last length BC.
we can use cosine rule to find the side BC.
Base on cosine rule
a² = b² + c² - 2bc cos A
a = BC
b = AC = 14
c = AB = 10
a² = 14² + 10² - 2 × 14 × 10 cos 51°
a² = 196 + 100 - 280 cos 51°
a² = 296 - 280 × 0.62932039105
a² = 296 - 176.209709494
a² = 119.790290506
square root both sides
a = √119.790290506
a = 10.9448750795
BC ≈ 10.94
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