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.

Respuesta :

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]

Answer Link
callmerory100
callmerory100 callmerory100
  • 17-02-2022

Answer:

342

Step-by-step explanation:

see image

Ver imagen callmerory100
Answer Link

Otras preguntas

How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2
How many neutrons does Uranium-238 have if it has 92 protons?
What does the phrase "¿Qué desayunas?" mean in English?
Which of these was a cause of the War of 1812? A. Canada was harboring slaves who escaped through the Underground Railroad. B. France was helping Native America
What effect might 2000 reindeer have on the island and its vegetation
How many lines of symmetry does a rhombus that is not a square have?
1kg costs £1.28. what is the cost of 250g
another company offers a rate of 0.05per minute how would you find the unit rate per hour
What time is it 15 minutes before 5:30
Which two European countries explored Canada between 1400-1700?