Question

In ΔJKL, k = 3.7 cm, ∠K=15° and ∠L=41°. Find the length of j, to the nearest 10th of a centimeter.

Answer 1

Answer: 11.9

Explanation:

nearest 10th you have to round it

Answer 2

The length of side j is 11.85cm

Explanation:

Given that

In triangle JKL,

`\angle k=15`

`\angle l=41`

Side k=LJ=3.7cm

To find side j=KL:

By using sine rule,

We can write as

`(SinK)/(LJ) = (SinL)/(JK) = (SinJ)/(KL) \\(Sin15)/(3.7) = (Sin41)/(JK) = (SinJ)/(KL)`

Using property of triangle,

`\angle k+\angle l+\angle j=180`

`15+41+\angle j=180`

`\angle j=124`

`(Sin15)/(3.7) = (Sin41)/(JK) = (Sin124)/(KL)\\(Sin15)/(3.7) = (Sin124)/(KL)\\KL=3.7(Sin124)/(Sin15)\\KL=3.7(0.8290)/(0.2588)\\KL=11.85cm`

Thus,

The length of side j is 11.85cm