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# URGENT HELP!! Thank you!!

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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
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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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