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Makovka662 [10]
3 years ago
8

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

Mathematics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:

L = 10.64°

Step-by-step 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

\dfrac{j}{SIn \ J} = \dfrac{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 = \sqrt{99.45}

j = 9.97

∴

\dfrac{9.97}{Sin \ 43} = \dfrac{2.7}{ Sin \ L}

{9.97 \times    Sin (L ) = (2.7 \times Sin \ 43)

=  Sin \ L = \dfrac{ (2.7 \times Sin \ 43)}{9.97 } \\ \\ =  Sin \ L = \dfrac{ (2.7 \times 0.6819)}{9.97 }  \\ \\  = Sin \ L = 0.18466 \\ \\  L = Sin^{-1} (0.18466) \\ \\  L = 10.64 ^0

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