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**+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! |

TheXSquaredFactor Sep 2, 2017