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Two angles of a triangle measure 30 and 45 degrees. If the side of the triangle opposite the 30-degree angle measures  \(12\) units, what is the area of the triangle? Express your answer as a decimal to the nearest tenth.

 Apr 15, 2022
 #1
avatar+2455 
+1

Draw an altitude that forms both a 45-45-90 triangle and a 30-60-90 triangle

 

Because we have a 30-60-90 triangle, the base of the triangle is \({12 \over \sqrt3} = {4 \sqrt3}\)

 

We also have a 45-45-90 triangle. We know 1 of the legs is 12, meaning the other leg must also be 12. 

 

Thus, the area of the triangle is \(12 \times (12 +4 \sqrt3) \div 2 \approx \color{brown}\boxed{113.6}\)

 

Here is a diagram: 

 

 Apr 15, 2022
edited by BuilderBoi  Apr 15, 2022
edited by BuilderBoi  Apr 15, 2022
 #2
avatar+124598 
+1

Another way

 

The third angle  =   105° 

 

Using the Law of Sines  the side opposite the 45°  angle = x =  can be found as 

 

x / sin 45 =  12 /sin 30

 

x =    12 sin 45 / sin 30  =    12 sqrt (2)

 

So....using the area of  a triangle knowing two sides  and  an included angle we have

 

A =  (1/2)  (12 sqrt 2 ) * ( 12 )  * [  sin (105°) ] =         

 

  Note :     {sin 105 = sin 75 }

 

72sqrt (2) *   sin (30 + 45)  =

 

72 sqrt (2)  [  sin 30 * cos 45  + sin 45  * cos 30 ] = 

 

72 sqrt (2) [ sqrt 2 / 4  +  sqrt 6 / 4 ]  =

 

72 (2/4)  + 18 sqrt (12)  =

 

36 + 18 * 2 sqrt (3)   =

 

36 + 36 sqrt (3) =  36 (1 + sqrt (3) )  units^2 ≈   98.4  units^2

 

cool cool cool

 Apr 15, 2022
edited by CPhill  Apr 15, 2022

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