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?
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! |