Given isosceles triangle, find sin (A-C)

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The discussion revolves around calculating sin(A-C) for an isosceles triangle ABC with sides AB = 10 cm and AC = BC = 13 cm. Participants explore the use of the angle difference formula sin(A-C) = sin A cos C - cos A sin C, questioning its necessity for the problem. They suggest utilizing the known side lengths to derive angles using trigonometric ratios without needing a calculator. Ultimately, one user confirms their calculations, finding that sin(A-C) equals approximately 0.376878, while acknowledging that using a calculator would simplify the process, which is not permitted in their test. The conversation highlights the application of trigonometric identities and the challenge of solving without computational tools.
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Homework Statement



Triangle ABC have side AB = 10 cm, AC=BC = 13cm, so sin (A-C) is...

2. Homework Equations

sin (A-C) = sin A cos C - cos A sin C

The Attempt at a Solution



I see that the triangle can be split into two right-angle triangles.
But, sin (A-C) ?? How to get that?[/B]
 
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terryds said:

Homework Statement



Triangle ABC have side AB = 10 cm, AC=BC = 13cm, so sin (A-C) is...

2. Homework Equations

sin (A-C) = sin A cos C - cos A sin C

The Attempt at a Solution



I see that the triangle can be split into two right-angle triangles.
But, sin (A-C) ?? How to get that?[/B]
Is there a sketch of this triangle?
 
SteamKing said:
Is there a sketch of this triangle?
Actually, there is no sketch of the triangle in the question.
But, here is my sketch
xzoifdlohaoniigd.png
 
terryds said:

Homework Statement



Triangle ABC have side AB = 10 cm, AC=BC = 13cm, so sin (A-C) is...

2. Homework Equations

sin (A-C) = sin A cos C - cos A sin C

The Attempt at a Solution



I see that the triangle can be split into two right-angle triangles.
But, sin (A-C) ?? How to get that?[/B]

use sohcahtoa

you know all the lengths, you can easily now work out all the angles.

http://www.mathwords.com/s/s_assets/s126.gif
 
Where did that equation come from?
terryds said:
2. Homework Equations sin (A-C) = sin A cos C - cos A sin C
 
NickAtNight said:
Where did that equation come from?
It's one of the well known angle sum/difference formulas.
 
SammyS said:
It's one of the well known angle sum/difference formulas.
True, but does he need that here?
 
NickAtNight said:
True, but does he need that here?
NickAtNight said:
True, but does he need that here?
Is it necessary? I don't know.

However, it can be used to solve the problem
 
SammyS said:
However, it can be used to solve the problem

Seems like the wrong equation to apply to me.

Edit: Never mind. Now I see. The next step makes it a simple multiplication, division and subtraction problem. That is a very elegant solution when applied that way. Cause we can apply the lengths directly !

We use the equations WW posted, but rather than calculate the angles, we eliminate the Sin's and Cos's. to get a simple math expression! No trig tables or Scientific calculator needed.

William White said:
use sohcahtoa

you know all the lengths, you can easily now work out all the angles.

http://www.mathwords.com/s/s_assets/s126.gif
 
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  • #10
NickAtNight said:
Seems like the wrong equation to apply to me.

Edit: Never mind. Now I see. The next step makes it a simple multiplication, division and subtraction problem. That is a very elegant solution when applied that way. Cause we can apply the lengths directly !

We use the equations WW posted, but rather than calculate the angles, we eliminate the Sin's and Cos's. to get a simple math expression! No trig tables or Scientific calculator needed.

So, should I use the formula sin (A-C) = sin A cos C - cos A sin C ??

sin A = 12/13
cos A = 5/13

sin (C/2) = 5/13
cos (C/2) = 12/13

Then,
sin (C) = 2 * sin (C/2) * cos (C/2) = 2 * (5/13) * (12/13) = 120/169
cos (C) = 1 - 2 sin^2(C/2) = 1 - 2 * 25/169 = 1 - 50/169 = 119/169

So,

sin (A - C) = sin A cos C - cos A sin C = 12/13 * 119/169 - 5/13*120/169 = 1428/2197 - 600/2197 = 828/2197

Do I get it right ?? Is there any simpler solution ?
 
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  • #11
Besides using a calculator with trig functions?
A = asin 12/13.
B=A
C = 180 - A - B

Dif = A-C

answer = Sin (dif)

edited. And edited
Sin (A) = (12/13) = 0.923. A = Asin (1.17) = 67.38 degrees. Same as B. C= 180-2A = 45.23
A-C = 67.38 - 45.23 = 22.14
Sin (22.14degrees) = sin (0.386 radians) = 0.376878. yes, your answer appears to be correct !
 
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  • #12
terryds said:
Do I get it right ?? Is there any simpler solution ?

By golly, it appears that you got it right ! (Took me a while to get the calculation right the other way)
 
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  • #13
NickAtNight said:
Besides using a calculator with trig functions?
A = asin 12/13.
B=A
C = 180 - A - B

Dif = A-C

answer = Sin (dif)

edited. And edited
Sin (A) = (12/13) = 0.923. A = Asin (1.17) = 67.38 degrees. Same as B. C= 180-2A = 45.23
A-C = 67.38 - 45.23 = 22.14
Sin (22.14degrees) = sin (0.386 radians) = 0.376878. yes, your answer appears to be correct !

Yap, using calculator is the simplest way, but it's not allowed in the test :(
 
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