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

Sagot :

Answer:

26

Step-by-step explanation:

JKL= ABC,  jkl=abc

1.) use law of sines to find k.

[tex]c^{2} =a^{2} +b^{2} -2abCosC[/tex]    

[tex]c=\sqrt{920^{2} +860^{2} -2(920)(860)cos(126) } = 1586.225515[/tex]

2.) use law of cosines to find L

[tex]\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}[/tex]  

plug in for b and c:  [tex]\frac{860}{sinB} =\frac{1586.225515}{sin126}[/tex]

cross multiply: [tex]\frac{1586.23sinB}{1586.23} =\frac{860sin126}{1586.23}[/tex]

[tex]sinB= 0.4386227612\\B=sin^-1(0.4386227612)= 26.01604086\\[/tex]

L=26

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