The problemFind side length using trig

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Homework Help Overview

The discussion revolves around a triangle problem involving side lengths and angles, specifically with given values of $AB = 20 cm$, $m∠A = 30°$, and $m∠C = 45°$. Participants are tasked with finding the length of side $BC$ in simplest radical form, utilizing trigonometric relationships.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between the sides and angles of the triangle, referencing 30-60-90 and 45-45-90 triangles. There is an attempt to apply sine ratios and the sine law, but confusion arises regarding the equality of sine values for different angles. Some participants express uncertainty about the side lengths corresponding to the trigonometric ratios.

Discussion Status

Several participants have provided insights and corrections regarding the application of trigonometric laws. There is an ongoing exploration of the relationships between the sides of the triangles, but no consensus has been reached on the correct approach or solution. Some guidance has been offered regarding the use of sine and cosine laws.

Contextual Notes

Participants note potential misunderstandings about the application of trigonometric ratios and the representation of mathematical expressions using LaTeX. There is also mention of the need for clarity in the use of symbols within the forum's formatting guidelines.

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Homework Statement



$AB = 20 cm$, $m∠A = 30°$ , and $m∠C = 45°$ . Express the number of centimeters in the length of $BC$ in simplest radical form.

Homework Equations


$sin A = sin C$

The Attempt at a Solution


$AB = 20, BC = x$

D is the point where this obtuse triangle separates into 2 right triangles

$BD/20 = sin A$
$AD/20 = cos A$

30-60-90 triangle
$1:2:\sqrt{3}$

BD is 10 according to this ratio which means that sin A is 1/2 and AD would be $20\sqrt{3}$

sin C is the same but for a 45-45-90 triangle instead.

45-45-90 triangle
$1:1:\sqrt{2}$

But here is where I am stuck. I am trying to find the side lengths of the 45-45-90 triangle with the trigonometric ratios being the same for both triangles but the angles being different so that I know the hypotenuse BC. But I don't know what side lengths will give me the trigonometric ratios being the same and the $1:2:\sqrt{3}$ and $1:1:\sqrt{2}$ side length ratios being true.
 
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caters said:

Homework Statement



$AB = 20 cm$, $m∠A = 30°$ , and $m∠C = 45°$ . Express the number of centimeters in the length of $BC$ in simplest radical form.

Homework Equations


$sin A = sin C$

The Attempt at a Solution


$AB = 20, BC = x$

D is the point where this obtuse triangle separates into 2 right triangles

$BD/20 = sin A$
$AD/20 = cos A$

30-60-90 triangle
$1:2:\sqrt{3}$

BD is 10 according to this ratio which means that sin A is 1/2 and AD would be $20\sqrt{3}$

sin C is the same but for a 45-45-90 triangle instead.

45-45-90 triangle
$1:1:\sqrt{2}$

But here is where I am stuck. I am trying to find the side lengths of the 45-45-90 triangle with the trigonometric ratios being the same for both triangles but the angles being different so that I know the hypotenuse BC. But I don't know what side lengths will give me the trigonometric ratios being the same and the $1:2:\sqrt{3}$ and $1:1:\sqrt{2}$ side length ratios being true.
Your result for the length of side AD is incorrect.
 
I second what SammyS says.
Once you know BD though, you should quickly know BC, since BC is the hypotenuse of the 45-45-90 triangle, right? You have already written the appropriate ratio for the length of a side to the hypotenuse of this triangle.
 
caters said:

Homework Statement



$AB = 20 cm$, $m∠A = 30°$ , and $m∠C = 45°$ . Express the number of centimeters in the length of $BC$ in simplest radical form.

Homework Equations


$sin A = sin C$
No, with A= 30 degrees and B= 45 degrees sin(A) is definitely not equal to sin(C)!
Perhaps you meant the sine law:
\frac{sin(A)}{BC}= \frac{sin(B)}{AC}= \frac{sin(C)}{AB}
The cosine law might also be useful:
(AB)^2= (AC)^2+ (BC)^2- 2(AC)(BC) cos(C)
and equivalent formulas for the other two angles.

3. The Attempt at a Solution
$AB = 20, BC = x$

D is the point where this obtuse triangle separates into 2 right triangles

$BD/20 = sin A$
$AD/20 = cos A$

30-60-90 triangle
$1:2:\sqrt{3}$

BD is 10 according to this ratio which means that sin A is 1/2 and AD would be $20\sqrt{3}$

sin C is the same but for a 45-45-90 triangle instead.

45-45-90 triangle
$1:1:\sqrt{2}$

But here is where I am stuck. I am trying to find the side lengths of the 45-45-90 triangle with the trigonometric ratios being the same for both triangles but the angles being different so that I know the hypotenuse BC. But I don't know what side lengths will give me the trigonometric ratios being the same and the $1:2:\sqrt{3}$ and $1:1:\sqrt{2}$ side length ratios being true.
 
I'm kind of new here... can someone please tell me what the dollar signs represent?
 
Saracen Rue said:
I'm kind of new here... can someone please tell me what the dollar signs represent?

They should be double dollar signs, and they're just a simpler alternative to the tags.
 
Saracen Rue said:
I'm kind of new here... can someone please tell me what the dollar signs represent?
To further explain:

The dollar signs (as well as the # sign ) are used as tags to enable using "LaTeX" for displaying mathematical expressions.

On this site, those should be doubled.

For instance, $20\sqrt{3}$ , should have been ## $ \$ 20\text{\sqrt } 3 ## ## $$ ## .

It would display $$20\sqrt{3}$$
 
Last edited:

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