In Triangle ABC, AB = 10. AC = 14, and angle A = 51°. Find the length of BC to the nearest hundredth

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

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

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )