Given isosceles triangle, find sin (A-C)

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Homework Help Overview

The problem involves an isosceles triangle ABC with sides AB = 10 cm and AC = BC = 13 cm, focusing on finding sin(A - C). Participants are discussing the application of trigonometric identities and relationships within the triangle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore splitting the triangle into right-angle triangles and question how to derive sin(A - C). There are discussions about the relevance of the angle sum/difference formula and whether it is necessary for this problem. Some suggest using known lengths to simplify calculations without relying on trigonometric tables or calculators.

Discussion Status

The discussion is active, with various participants offering insights and alternative methods for approaching the problem. Some express uncertainty about the appropriateness of certain formulas, while others suggest that a direct application of the triangle's dimensions may lead to a simpler solution.

Contextual Notes

Participants note that the use of calculators is not permitted in the test, which influences the methods being considered. There is also a lack of a visual sketch of the triangle, which some participants indicate could aid in understanding the problem better.

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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
 
Last edited:
  • #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 !
 
Last edited:
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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)
 
Last edited:
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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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