Answer :
Answer:
In ΔJKL, k = 430 cm, l = 590 cm and ∠J=117°. Find the length of j, to the nearest centimeter.
Step-by-step explanation:
\text{S.A.S.}\rightarrow \text{Law of Cosines}
S.A.S.→Law of Cosines
a^2=b^2+c^2-2bc\cos A
a
2
=b
2
+c
2
−2bccosA
From reference sheet.
j^2 = 430^2+590^2-2(430)(590)\cos 117
j
2
=430
2
+590
2
−2(430)(590)cos117
Plug in values.
j^2 = 184900+348100-2(430)(590)(-0.45399)
j
2
=184900+348100−2(430)(590)(−0.45399)
Square and find cosine.
j^2 = 184900+348100+230354.77957
j
2
=184900+348100+230354.77957
Multiply.
j^2 = 763354.77957
j
2
=763354.77957
Add.
j=\sqrt{763354.77957} \approx873.702 \approx874
j=
763354.77957
≈873.702≈874
Square root and round.