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

 Jun 2, 2024
 #1
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         B

    

    6         8

 

A                   C 30

 

Law pf Cosines

 

AB^2  = AC^2 + BC^2  - 2(AC * BC) cos (30)

 

6^2 = AC^2 + 8^2  -  2 (AC * 8C) sqrt (3)/2

 

36  - 64 = AC^2   - 8sqrt (3) AC

 

-28 = AC^2 - 8sqrt (3) AC

 

AC^2  - 8sqrt 3 AC  + 28 =  0       solving for AC

 

AC   ≈ 2.46    or  AC ≈  11.4

 

So..because pf the SSA situation.....two possible areas result

 

[ ABC] =   (1/2) (2.46) (8) (1/2)  ≈ 4.92

[ABC ]  =  (1/2) (11.4) (8)(1/2)  ≈  22.8

 

cool cool cool

 Jun 2, 2024
edited by CPhill  Jun 2, 2024

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