# Thread: Sine Rule Problem; WHY 2 Answers!??

1. ## Sine Rule Problem; WHY 2 Answers!??

Alright, so...

I put sinx/70=sin40/46. so i got the answer x=78
but there are two answers? 102 and 78?

Someone explained to me that you subtract 78 from 180 for the second answer. But why do i do that? Do i ALWAYS do that? Or is it only in a certain case? And what case is it? How would I know when to subtract it from 180 for a second answer?

Sorry about the silly questions! I haven't really learned this stuff but I'm trying to study a little in advance during the summer.

2. $sinx=\frac{70sin40^o}{46}$

Now, you have $0

Example: $siny=\frac{1}{2}$(say) has $infinite$ solutions....which one you will choose???

It depends on the range of your answer.

You have range $0, and there are two solutions in that range for your question.

3. Because this is an SSA case (side-side-angle) case. It's ambiguous; if you're given a triangle with SSA, you don't know how many triangles are possible. It could be 0, 1, or 2. In this case, it's two. See diagram below. I rearranged it so it's easier to see.

The information given could result in triangle ABC with x = 78°, as you figured out. But the triangle could also be drawn as in triangle ABD, with x = 102°.

How can you tell how many triangles are possible? You can draw the height. But you have to draw the correct height. You start at the vertex between the two known sides and drop a perpendicular to the unknown side. In your diagram, you would start at the lower left vertex and drop a perpendicular to the opposite side. You can then find the length of this height with a simple sine ratio. In our case:

$\sin 40^{\circ} = \frac{h}{70}$
$h = 70\sin 40^{\circ} \approx 45$

Now compare the lengths of the height and the two known sides. The height is the shortest, followed by the side opposite the known angle, and the adjacent side is the longest. Therefore we have two triangles.

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4. Here are some more diagrams that illustrate the possible cases for triangle with SSA given.

The known angle is θ (angle B), the side adjacent to the known angle is a, and the side opposite the known angle is b.

In the first diagram, h < a < b. There is only one triangle possible.
In the second diagram, h < b < a. There are two triangles possible: one is triangle A1BC, and the other is triangle A2BC.
And in the third diagram, b < h < a. There are no triangles possible.

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