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8 days ago

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

Mathematics

1 answer:

Svet_ta [11.3K]8 days ago

8 0

Response:

The measurement of side j is 11.85cm.

Step-by-step clarification:

Given the situation

In triangle JKL,

$\angle k=15$

$\angle l=41$

Side k=LJ=3.7cm

To determine side j=KL:

Utilizing the sine rule

We can express it as

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

Using the triangle's property

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

$15+41+\angle j=180$

$\angle j=124$

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

Hence,

The measurement of side j is 11.85cm.

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