imaniconwayvega imaniconwayvega
  • 24-05-2021
  • Mathematics
contestada

In ΔJKL, j = 27 cm, k = 82 cm and ∠L=162°. Find the area of ΔJKL, to the nearest square centimeter.

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Corsaquix
Corsaquix Corsaquix
  • 24-05-2021

Answer:

[tex]684\:\mathrm{cm^2}[/tex]

Step-by-step explanation:

The area of any triangle is equal to [tex]A=\frac{1}{2}\cdot a\cdot b\cdot \sin C[/tex], where [tex]a[/tex] and [tex]b[/tex] are two sides of a triangle and [tex]C[/tex] is the angle between them.

Plugging in given values, we have:

[tex]A=\frac{1}{2}\cdot 27\cdot 82\cdot \sin 162^{\circ}=\boxed{684\:\mathrm{cm^2}}[/tex]

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callmerory100
callmerory100 callmerory100
  • 17-02-2022

Answer:

342

Step-by-step explanation:

see image

Ver imagen callmerory100
Answer Link

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