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avatar+388 

Find the area of triangle $ABC$ if $AB = 6,$ $BC = 8,$ and $\angle ACB = 30^\circ$.

 Jun 3, 2024
 #1
avatar+104 
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Hey, I just wanna know why are you using the same pfp and name as me...

 Jun 4, 2024
 #2
avatar+129487 
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         B

    

     6           8

 

A                  C 30

 

We have a SSA  situation......there may be 1 triangle formed, 2 triangles formed or no triangles formed

 

Law of Sines

 

sin A / 8  =  sin 30 / 6

 

sin A / 8 = (1/2) / 6

 

sinA = (8/6) (1/2) = 2/3

 

arcsin (2/3) ≈  41.81°    or    180 - 41.81  = 138.19

 

So   angle B  =  180  - 30  - 41.81   =  108.19°

or    angle B =  180 - 30 - 138.19  =    11.81°

 

If B =  108.19°   area of ABC =  (1/2) (8)(6)sin (108.19°)     ≈  22.8

 

If B  =  11.81°  area of ABC  = (1/2) (8)(6) sin ( 11.81°)  ≈  4.9

 

cool cool cool

 Jun 4, 2024

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