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:

The length of side j is 11.85cm

Step-by-step 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

$\frac{SinK}{LJ} = \frac{SinL}{JK} = \frac{SinJ}{KL} \\\frac{Sin15}{3.7} = \frac{Sin41}{JK} = \frac{SinJ}{KL}$

Using property of triangle,

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

$15+41+\angle j=180$

$\angle j=124$

$\frac{Sin15}{3.7} = \frac{Sin41}{JK} = \frac{Sin124}{KL}\\\frac{Sin15}{3.7} = \frac{Sin124}{KL}\\KL=3.7\frac{Sin124}{Sin15}\\KL=3.7\frac{0.8290}{0.2588}\\KL=11.85cm$

Thus,

The length of side j is 11.85cm

Answer 2

Answer: 11.9

Step-by-step explanation:

nearest 10th you have to round it