+0  
 
0
30
1
avatar

Find the area of triangle ABC.

 

 Apr 27, 2022
 #1
avatar+122385 
+1

We have a SSA  situation.....I'm assuming the  greater possible value  for BC

 

Use Law of Cosines to find   BC

 

9^2 = 10^2 + BC^2  -2 (10) (BC)cos 60

 

81 - 100  =  BC^2  - 10BC

 

-19  =  BC^2  - 10BC

 

BC^2  - 10BC  + 19   =  0

 

BC^2  -10BC =  -19                     complete the square on BC

 

BC^2  - 10BC  + 25  =  -19  + 25

 

(BC - 5)^2  =   6                take the positive root

 

BC - 5  =   sqrt 6

 

BC  =  5 +  sqrt 6

 

Area =  (1/2)(AC)(BC) sin 60  = (1/2) (5 +sqrt 6) (10) (sin 60)  =  5 (5 + sqrt 6) ( sqrt (3) / 2 )    ≈

 

32.26 units^2

 

 

cool cool cool

 Apr 27, 2022
edited by CPhill  Apr 27, 2022

28 Online Users