Question

In ΔJKL, k = 9.6 cm, l = 2.7 cm and ∠J=43°. Find ∠L, to the nearest 10th of a degree.

Answer 1

L = 10.64°

Explanation:

From the given information:

In triangle JKL;

line k = 9.6 cm

line l = 2.7 cm; &

angle J = 43°

we are to find angle L = ???

We can use the sine rule to determine angle L:

i.e

`(j)/(SIn \ J) = (l)/( SIn \ L)`

Using Pythagoras rule to find j

i,e

j² = k² + l²

j² = 9.6²+ 2.7²

j² = 92.16 + 7.29

j² = 99.45

`j = √(99.45)`

j = 9.97

∴

`(9.97)/(Sin \ 43) = (2.7)/( Sin \ L)`

`{9.97 * Sin (L ) = (2.7 * Sin \ 43)`

`= Sin \ L = ( (2.7 * Sin \ 43))/(9.97 ) \\ \\ = Sin \ L = ( (2.7 * 0.6819))/(9.97 ) \\ \\ = Sin \ L = 0.18466 \\ \\ L = Sin^(-1) (0.18466) \\ \\ L = 10.64 ^0`