Question

In ΔJKL, l = 860 inches, j = 920 inches and ∠K=126°. Find ∠L, to the nearest degree.

Answer 1

26

Explanation:

JKL= ABC, jkl=abc

1.) use law of sines to find k.

`c^(2) =a^(2) +b^(2) -2abCosC`

`c=\sqrt{920^(2) +860^(2) -2(920)(860)cos(126) } = 1586.225515`

2.) use law of cosines to find L

`(a)/(sinA) =(b)/(sinB) =(c)/(sinC)`

plug in for b and c:
`(860)/(sinB) =(1586.225515)/(sin126)`

cross multiply:
`(1586.23sinB)/(1586.23) =(860sin126)/(1586.23)`

`sinB= 0.4386227612\\B=sin^-1(0.4386227612)= 26.01604086\\`

L=26