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enter image description here

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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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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  • $\begingroup$ But shouldn't the lowercase letter be opposite of the Upper case angle? If i do the way you described, i would have to re-arrange the whole thing again? Is that how i do it? and is their a way i can do the way you described in the first try $\endgroup$ – MATH ASKER Dec 4 '15 at 1:03
  • $\begingroup$ and also my teacher really didn't say anything was wrong with my diagram ..except the fact that the lower case letter weren't opposite of the larger one $\endgroup$ – MATH ASKER Dec 4 '15 at 1:04
  • $\begingroup$ than what would have happened if A was instead of C in the diagram above? is their a proper way to label them and solve? $\endgroup$ – MATH ASKER Dec 4 '15 at 1:09
  • $\begingroup$ The letters you use to label the sides don't matter, except as a mnemonic device to help you keep things straight. The important point is that it is impossible to have a triangle shaped with the measurements you have written on your triangle. $\endgroup$ – mweiss Dec 4 '15 at 1:12
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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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  • $\begingroup$ that would be about 40.0 and it does't match with other answers, $\endgroup$ – MATH ASKER Dec 4 '15 at 0:36
  • $\begingroup$ Ah, I think you have angles $A$ and $B$ backwards. If $B>A, b>a$, which isn't the case in your picture. I've edited my post such that angle $A$ is $63$, assuming triangle $ABC$ exists. $\endgroup$ – Lanier Freeman Dec 4 '15 at 1:04
  • $\begingroup$ mweiss identified this issue in his response. $\endgroup$ – Lanier Freeman Dec 4 '15 at 1:07
  • $\begingroup$ than what would have happened if A was instead of C in the diagram above? is their a proper way to label them and solve? $\endgroup$ – MATH ASKER Dec 4 '15 at 1:09
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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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  • $\begingroup$ I tried doing it but each time i chose different sides i got different answers $\endgroup$ – MATH ASKER Dec 3 '15 at 23:28
  • $\begingroup$ @MATHASKER Please show us your work. It's hard to explain what is going wrong without seeing it. $\endgroup$ – mweiss Dec 4 '15 at 0:39

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