Question

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

Answer 1

Answer:

26

Step-by-step 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

$\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}$  

plug in for b and c:  $\frac{860}{sinB} =\frac{1586.225515}{sin126}$

cross multiply: $\frac{1586.23sinB}{1586.23} =\frac{860sin126}{1586.23}$

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

L=26