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To find the area of this triangle we use $$ \text{area} = \frac 1 2 \cdot a \cdot b \cdot \sin(C)$$ How do i know when to use the sin of anything like lets say The triangle is $$ \angle DEF $$ how do i know which one i would take the sin of to find the area? like the general idea? I got Different answers every time i chose two sides and an angle between them

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3 Answers

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The reason you are getting different answers when you try to calculate this different ways is because your diagram shows an impossible triangle. If you have sides of length 9, 29.1, and 32.05, then there will be a 63.06° angle -- but it will be between the two longest sides of the triangle. Likewise, the 16° angle will be across from the shortest side of the triangle.

Try re-drawing your triangle and relabeling the sides and angles so that the longest side is across from the largest angle and the shortest side is across from the shortest angle. Then your calculations should work.

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Base $c$ equals $32.05.$

Calculate height $h$ using $9\sin(63)$.

Your area should be $\frac{1}{2}*32.05*9\sin(63)$

I wouldn't bother with taking the sine of $C$ for multiple reasons. First, you have an obtuse triangle, so it's best to split the triangle at the large angle. second, you're complicating things by taking the sine of $C$ because you'll have to reorient your triangle.

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To use the formula $A = \frac{1}{2} ab \sin(C)$ you need to know two side-lengths and the angle between them. If you know all three side-lengths and all three angles, as in the diagram, then you can just pick any two sides you want; then take the angle in between them.

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