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avatar+326 

There are two triangles satisfying \(\angle BCA = 60^{\circ}, AB = 9\) and \(AC = 10\) shown below:

 

 

The possible values of BC are a1 and a2, as shown in the picture, with a1

 

a1, a2

 

in that order?

 Apr 14, 2022
 #1
avatar+122390 
+1

We have an example of the ambiguous case = SSA

 

Using the Law of Sines

 

sin B / 10  = sin C / 9

sin B /10 = [sqrt (3) /2] / 9

sin B = 10 sqrt (3) /  18 = (5/9)sqrt 3 =  

arcsin [ (5/9)sqrt 3 ] = B  ≈  74.2°  or    B  ≈ 105.8°

 

So

 

So , in one case, C =  180  - 60  - 74.2   ≈ 45.8°

 

So...let's figure a2 first

 

a2 / sin (45.8)  = 9 /sin C

a2 / sin (45.8) =  9 / [ sqrt (3) / 2 ]

a2 =sin (45.8) * 18 / sqrt 3  ≈  7.45 

 

And in the second case

C = 180 -60 - 105.8   ≈  14.2°

 

So....by the Law of Sines again we can find a1 as

a1 / sin (14.2) = 18/sqrt 3

a1=  sin (14.2) * 18 /sqrt 3 ≈  2.55

 

{a1 , a2}  = {2.55 , 7.45}

 

 

cool cool cool

 Apr 14, 2022
 #2
avatar+326 
+1

Thanks once again CPhill!!!

-Vinculum

 Apr 14, 2022
 #3
avatar+122 
-1

Chris is a genius

coolcoolcool

 Apr 14, 2022

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