# Urgent Trig Help law of sines

• Nov 25th 2006, 02:53 PM
Lyrixa
Urgent Trig Help law of sines
Hi!

I am learning the Law of Sines, and have no idea what to do. I have notes from my class, but the whole multiple solutions thing is throwing me. I do, however, understand the basics about the law of Sines.

Here is a problem I don't understand- it is an Ambiguous case:

Determine the number of possible solutions. If a solution exists, solve the triangle.
Triangle ABC
C=17 degrees a= 10 units c= 11 units

How would I go about finding <A and side B so that I can check to see how many solutions?

Thanks,

Lyrixa
• Nov 25th 2006, 03:46 PM
Ranger SVO
Lets pretend for a minute that we have a triangle with three sides measuring 5-inches, 8-inches and 2-inches.

No triangle exists for these measurements. Why?

Because side A + side B must be greater than side C ( I always pick the longest side for C
• Nov 25th 2006, 03:58 PM
ThePerfectHacker
Quote:

Originally Posted by Lyrixa
Hi!

I am learning the Law of Sines, and have no idea what to do. I have notes from my class, but the whole multiple solutions thing is throwing me. I do, however, understand the basics about the law of Sines.

Here is a problem I don't understand- it is an Ambiguous case:

Determine the number of possible solutions. If a solution exists, solve the triangle.
Triangle ABC
C=17 degrees a= 10 units c= 11 units

How would I go about finding <A and side B so that I can check to see how many solutions?

Thanks,

Lyrixa

We know by the law of sines,
$\frac{11}{\sin 17^o}=\frac{10}{\sin A}$
Cross multiply,
$11\sin A=10\sin 17^o\approx 2.92$
Divide both sides by 11,
$\sin A=.265$
Using the inverse sine function on calculator (that is $\boxed{\sin^{-1}}$).
You get,
$A\approx 15^o$
But wait!
There is still another possible angle (in Second Quadrant, because sine is positive in first and second).
So,
$A=180^o -15^o=165^o$
So there are 2 possible triangles.
And those are the angles that angle $A$ can have.
• Nov 25th 2006, 03:59 PM
Lyrixa
Quote:

Originally Posted by Ranger SVO
Lets pretend for a minute that we have a triangle with three sides measuring 5-inches, 8-inches and 2-inches.

No triangle exists for these measurements. Why?

Because side A + side B must be greater than side C ( I always pick the longest side for C

Okay... but if you were only given two sides, how would you find the value of the third so that you can see if sideA+sideB would be greater than sideC? Like if you were only given 5 inches and 2 inches?
• Nov 25th 2006, 04:03 PM
Lyrixa
Quote:

Originally Posted by ThePerfectHacker
We know by the law of sines,
$\frac{11}{\sin 17^o}=\frac{10}{\sin A}$
Cross multiply,
$11\sin A=10\sin 17^o\approx 2.92$
Divide both sides by 11,
$\sin A=.265$
Using the inverse sine function on calculator (that is $\boxed{\sin^{-1}}$).
You get,
$A\approx 15^o$
But wait!
There is still another possible angle (in Second Quadrant, because sine is positive in first and second).
So,
$A=180^o -15^o=165^o$
So there are 2 possible triangles.
And those are the angles that angle $A$ can have.

Okay, figuring the value of A1 now makes sense, but the only way you know that there is another triangle is because sine is positive in both the 1st and 2nd Quadrants? Wouldn't that make every triangle have 2 solutions?
• Nov 25th 2006, 04:06 PM
ThePerfectHacker
Quote:

Originally Posted by Lyrixa
Okay, figuring the value of A1 now makes sense, but the only way you know that there is another triangle is because sine is positive in both the 1st and 2nd Quadrants? Wouldn't that make every triangle have 2 solutions?

No!

Say you end up with the following, after some algebra,
$\sin A=2$
There is no solution!
Because the largest value of sine is 1.
So there is no triangle.

WAIT!!!
UPDATE

I made a mistake so the value of $A$ is:
$15,165$
But it cannot be $165$
Because the other angle you are given is,
$17$
So when you add them there is no way you can get $180$ degree like in a triangle.
So it can only have 1 angle.
• Nov 25th 2006, 04:19 PM
Lyrixa
Quote:

Originally Posted by ThePerfectHacker
No!

Say you end up with the following, after some algebra,
$\sin A=2$
There is no solution!
Because the largest value of sine is 1.
So there is no triangle.

WAIT!!!
UPDATE

I made a mistake so the value of $A$ is:
$15,165$
But it cannot be $165$
Because the other angle you are given is,
$17$
So when you add them there is no way you can get $180$ degree like in a triangle.
So it can only have 1 angle.

Okay... That makes a lot more sense... so on the second triangle, $180 -angleA2 -angleB = angleC$, and there is only a second triangle if $angleC$ is positive?
• Nov 25th 2006, 04:23 PM
ThePerfectHacker
Quote:

Originally Posted by Lyrixa
Okay... That makes a lot more sense... so on the second triangle, $180 -angleA2 -angleB = angleC$, and there is only a second triangle if $angleC$ is positive?

That is how you use it to find the third angle.
But if you get the angle as a negative answer like here it means we have a contradiction, that there is no such triangle with such a side.
• Nov 25th 2006, 04:25 PM
Lyrixa
Okay... I think I understand all of this now. Thank you so much for your help!!!!!

Lyrixa :D
• Nov 25th 2006, 04:36 PM
Ranger SVO
Only one triangle is possible.

I hope someone can give you a better explination
Sin(C)/c = Sin(A)/a so we have Sin(17)/11 = Sin(A)/10

We should get (10*Sin(17))/11 = Sin(A)
Sin(A) should equal .266 which will get us an angle of about 15.414-degrees

Now look at the picture, Line A and Line C are fixed at 10 and 11. It is not possible for me to move line C and make another triangle.
• Nov 25th 2006, 04:48 PM
Lyrixa
Quote:

Originally Posted by Ranger SVO
Only one triangle is possible.

I hope someone can give you a better explination
Sin(C)/c = Sin(A)/a so we have Sin(17)/11 = Sin(A)/10

We should get (10*Sin(17))/11 = Sin(A)
Sin(A) should equal .266 which will get us an angle of about 15.414-degrees

Now look at the picture, Line A and Line C are fixed at 10 and 11. It is not possible for me to move line C and make another triangle.

I see that... You can't move line C because it is fixed at 11 and would be too short!! Thanks for using the picture, I guess I lean more toward being a visual learner.

~Lyrixa
• Nov 25th 2006, 05:02 PM
Ranger SVO
I cannot stop drawing pictures. In the sketch below I think you can see why two triangles are possible. Angle C = 26.6-degrees with Side C = 4.5-inches and side A is 9-inches. I can move side C and maintain its fixed length.

Angle A can equal 63.2-degrees or 180-62.8 = 117.2-degrees
• Nov 25th 2006, 07:01 PM
Lyrixa
Quote:

Originally Posted by Ranger SVO
I cannot stop drawing pictures. In the sketch below I think you can see why two triangles are possible. Angle C = 26.6-degrees with Side C = 4.5-inches and side A is 9-inches. I can move side C and maintain its fixed length.

Angle A can equal 63.2-degrees or 180-62.8 = 117.2-degrees

Thank you sooo much!! Because of you and ThePerfectHacker, I was able to finish the problems on the Law of Cosines with ease!

Thanks Again!!

Lyrixa

P.S. The pictures are awesome, and a big help!
• Nov 25th 2006, 08:08 PM
Soroban
Hello, Lyrixa!

Quote:

I am learning the Law of Sines, and have no idea what to do.
I have notes from my class, but the whole multiple solutions thing is throwing me.
Don't blame the multiple solutions . . . You can still solve the triangle, right?

I do, however, understand the basics about the Law of Sines. You do?

Determine the number of possible solutions. If a solution exists, solve the triangle.

Triangle ABC: . $C = 17^o,\;a = 10,\;c= 11$

How would I go about finding <A and side B?
But you said you understand the basics of the Law of Sines.
Why can't you find angle A and side B?

We have: . $\boxed{\begin{array}{ccc}A\:= \\B\:=\\C\:=\end{array}
\begin{array}{ccc}a\:=\\b\:=\\c\:=\end{array}

We want angle A.

Law of Sines: . $\frac{\sin A}{a} \:= \:\frac{\sin C}{c}\quad\Rightarrow\quad \sin A \:= \:\frac{a\sin C}{c}$

So we have: . $\sin A \:= \:\frac{10\sin17^o}{11} \:\approx\:0.2658$

Then: . $A \:=\:\sin^{-1}(0.2658) \:\approx\:15.4^o$ . . . or $164.6^o$ **

If $A = 15.4^o$, then $B \,=\,180^o - 15.4^o - 17^o \:=\:147.6^o$

And we have: . $\boxed{\begin{array}{ccc}A\:=\ \\ B\;= \\ C\:=\end{array}
\begin{array}{ccc}15.4^o \\ 147.6^o \\ 17^o\end{array}
\begin{array}{ccc} a\:=\\b\:=\\c\:=\end{array}

Then: . $\frac{b}{\sin B} \:=\:\frac{c}{\sin C}\quad\Rightarrow\quad b \:=\:\frac{c\sin B}{\sin C}$

So we have: . $b\:=\:\frac{11\sin147.6^o}{\sin17^o} \:\approx\:20.2$

And we have solved the triangle: . $\boxed{\begin{array}{ccc}A\:=\ \\ B\;= \\ C\:=\end{array}
\begin{array}{ccc}15.4^o \\ 147.6^o \\ 17^o\end{array}
\begin{array}{ccc} a\:=\\b\:=\\c\:=\end{array}
\begin{array}{ccc}10 \\ 20.2\\11\end{array}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Every time you use the $\boxed{\sin^{-1}}$ key, there are two possible answers.

One is the answer on your calculator screen, always an acute angle.
. . The other is its supplement (subtract from 180°)

We must check to see if the other angle is a possible answer.
. . In this case, no.

If $A = 164.6^o$, our chart would looks like this:

. . $\boxed{\begin{array}{ccc}A\:=\ \\ B\;= \\ C\:=\end{array}
We see that $A + C \:=\:181.6^o$ ... which is impossible.