Solve tiangle

Dec 2009
54
0
Solve triangle in which is

$$ b+c=20,\ a=5\sqrt 2,\ \gamma=135°$$

($2$ solutions)
 
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Plato

MHF Helper
Aug 2006
22,475
8,643
Solve triangle in which is
$$ b+c=20,\ a=5\sqrt 2,\ \gamma=135°$$
($2$ solutions)
Normally the word solve is applied to an equation. I don't see any, do you?
So you need to explain and show some effort. This is not a homework service.
 
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Feb 2015
2,255
510
Ottawa Ontario
u = alpha, v = beta, w = gamma
w = 135, v = 180 - 135 - u = 45 - u

5sqrt(2) / SIN(u) = (20-b) / SIN(135) = b / SIN(45 - u)

Above is a hint.
Sorry, homework not done here...

Not sure what you're solving for...so solve for whatever turns you on!!
 
Dec 2009
54
0
Solve triangle means find basic elements of triangle
$$ a,\ b,\ c,\ \alpha,\ \beta,\ \gamma.$$
I have solutions which demand picture which I don't know how to draw. And, please somebody to tell me where to post problems like challenges.

If $D-C-A,\ and\ BD=h_b$ then $BD=DC=5$ and from $\Delta ABD$ with Pithagoras law we have

$$(20-b)^2=5^2+(b+5)^2$$

which gives $b=7$. Te rest is easy, but this give me just one solution.
 
Last edited:
Feb 2015
2,255
510
Ottawa Ontario
Solve triangle means find basic elements of triangle
$$ a,\ b,\ c,\ \alpha,\ \beta,\ \gamma.$$
I have solutions which demand picture which I don't know how to draw. And, please somebody to tell me where to post problems like challenges.

If $D-C-A,\ and\ BD=h_b$ then $BD=DC=5$ and from $\Delta ABD$ with Pithagoras law we have

$$(20-b)^2=5^2+(b+5)^2$$

which gives $b=7$. Te rest is easy, but this give me just one solution.
No idea what you're doing...
You gave one of the angles as 135: so IMPOSSIBLE for triangle to be a right triangle...

Are you a student attending math classes?
 
Dec 2009
54
0
Ha, ha, ha, I'm professor of math from Serbia. Draw the picture, all is clear.
 

Plato

MHF Helper
Aug 2006
22,475
8,643
What the hell does that mean????
Can you at least describe what the diagram would look like?
Denis, $D-C-A$ is standard notation is axiomatic geometry. It means that $D,~C,~\&~A$ are colinear points and $C$ is between $D~\&~A$.
But I'll be damn if I know what this professor troll means by it.
 
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Walagaster

MHF Helper
Apr 2018
223
144
Tempe, AZ
It's always interesting when the first problem is trying to decipher what the poster is really trying to do. I have attached picture below (not necessarily to scale) of my best guess of what the professor is looking for. I'm guessing the $135^\circ$ may be interior or exterior to the triangle giving an alternate angle $A'$ and corresponding adjacent sides $b'$ and $c'$ to give two triangles.
Here's a picture (click on it to expand):
 

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Feb 2015
2,255
510
Ottawa Ontario
Only someone from Tempe AZ could figure that out :)

Also looks like a 7-24-25 right triangle is involved...right?
 
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