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Help would be greatly appreciated!! Thanks so much!!

How do I calculate the area of an isosceles triangle in which 2 of its angles are equal to 75 degrees (the third angle would be 30 degrees) and 2 of its sides have a length of 1?

Guest Sep 2, 2017
 #1
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Here's a diagram to reference as I solve. I made it myself.

 

 

Despite that it is not noted on my diagram, \(m\angle B=30^{\circ}\), according to the given info. We already have enough information to solve for the area of the triangle.

 

In a triangle, if the length of 2 sides and measure of the included angle are known, then we can use the following formula to determine the area of the triangle. It is the following:

 

\(A_{\triangle}=\frac{1}{2}ab*\sin C\)

 

a = 1 side length of triangle

b = Another side length of the triangle 

C = measure of the included angle

 

Now that we know this formula, just substitute into it.

 

\(A_{\triangle}=\frac{1}{2}ab*\sin C\)Plug in the appropriate values for a, b, and C.
\(A_{\triangle}=\frac{1}{2}*1*1*\sin (30)\)You may have remembered that \(\sin(30)=\frac{1}{2}\).
\(A_{\triangle}=\frac{1}{2}*1*1*\frac{1}{2}\)Of course, 1 mulriplied by a number is itself.
\(A_{\triangle}=\frac{1}{2}*\frac{1}{2}\)Simplify this.
\(A_{\triangle}=\frac{1}{4}units^2=0.25units^2\)Of course, do not forget to include the units!
  
TheXSquaredFactor  Sep 2, 2017

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