"We're given a triangle $ABC$, with side $AC = 5$, $\angle CAB = 30^\circ, \angle ACB=45^\circ$. The triangle revolves around side $AC$. Find the surface area of the resulting figure."
I've solved these kinds of problems with right triangles before, but I can't seem to grasp what the result of revolving an obtuse triangle is.
Here's the given picture:
The picture is kind of confusing to me. From what I understand, the radius of the circle would be the side $OC = OB$. I can find the sides of the triangle using the Law of Sines, but the main problem is i'm not understanding the picture.
You said that you have done similar problems with right triangles. Well, then you can break the given obtuse triangle into two right triangles $\Delta ABO$ and $\Delta CBO$ and finally add the areas (curved surface areas) of the two cones which you get after revolving these two right triangles.
Solution- By Sine rule we get, $AB=\frac {5\sqrt{2}}{\sqrt{3}+1}$ and $CB=\frac {10}{\sqrt{3}+1}.$We can also find $AO$ and $CO$ as they are equal to $AB\cos (30)$ and $CB\cos (45) $respectively. $$$$ Total area=area due to $\Delta ABO$ + area due to $\Delta CBO$=$\pi(AO)(AB)+\pi(CO)(CB)$$$$$ Since all the values are Known we can find the total area.