69,125 views
6 votes
6 votes
In ΔJKL, k = 430 cm, l = 590 cm and ∠J=117°. Find the length of j, to the nearest centimeter

User Shafique
by
2.9k points

1 Answer

24 votes
24 votes

Answer:

j²=k²+l²-2klCosJ

j²=430²+590²-2(430)(590)Cos117

j=874cm

Explanation:

Use the formula of triangles.

make sure to square root at the second last step where j²=763 354,7797

therefore your answer will be 874cm if you round off correctly to the nearest centimeter.

User Tom Morris
by
2.6k points