Need help to solve a geometry problem

[shan86]

In above diagram X, Y and R are given. I need to find a and b angles. Please let me know how to do that. Thanks

 

[Harry_the_cat]

If the angle between side x and side y is a right angle, you could use Pythagoras' theorem to find the length of the other dotted line. Then use some basic trig to find angle b.

 

[mmm4444bot]

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[lex]

We need to know is the line marked x parallel to the line with the arrow at the bottom, and is the line marked y perpendicular to the line marked x.

 

[shan86]

If the angle between side x and side y is a right angle, you could use Pythagoras' theorem to find the length of the other dotted line. Then use some basic trig to find angle b.

Hi, thanks for the reply. Yes, x and y is a right angle. And I can find the length of the hypotenuse. But then, I don't know how to find those angles. Please help me

 

[shan86]

We need to know is the line marked x parallel to the line with the arrow at the bottom, and is the line marked y perpendicular to the line marked x.

Hi, thanks for the reply. Yes, x is parallel to the arrow at the bottom. And x and y forms a right angle. Please let me know how to find those angles.

 

[shan86]

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Welcome to our tutoring boards! :) This page summarizes main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to help...

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Ok, sorry I missed that. I'm a software developer and I need to work on a project that involves 2D shapes. This is stripped down version of the problem that I'm facing right now. Basically, I need to draw this shape, based on the X, Y and R, which are user inputs.

 

[lex]

Basically you need to find the co-ordinates of the corner of the shape (at the right angle).
Altering your notation slightly:

you simply need to solve the simultaneous quadratic equations in x and y:
1. [imath](x+u)^2+y^2=r^2[/imath]
2. [imath]x^2+(y+v)^2=r^2[/imath]

This yields a solution [imath]x=\dfrac{-u+\sqrt{u^2-4\alpha}}{2}, y=\sqrt{r^2-(x+u)^2}[/imath]
where [imath]\alpha=\dfrac{u^2+v^2}{4}-\dfrac{v^2r^2}{u^2+v^2}[/imath]

(Depending on whether u and/or v are less than r, there may be a corresponding solution in each of the 4 quadrants, in 2 quadrants, in only 1 quadrant or no solution if u, v are 'too large' relative to r).

 

[Dr.Peterson]

If you need to find the angles rather than coordinates, I would do this:



Using the right triangle with legs x and y, find the diagonal, and also find the angle marked in red. This is congruent to the red angle at the origin, because the sides of the angles are perpendicular.

Then using the right triangle with the red line as a leg, find the angle marked in blue. From those two angles, you can find a and b.

 

[lex]

Yes, much nicer!

 

[shan86]

Basically you need to find the co-ordinates of the corner of the shape (at the right angle).
Altering your notation slightly:
View attachment 28857
you simply need to solve the simultaneous quadratic equations in x and y:
1. [imath](x+u)^2+y^2=r^2[/imath]
2. [imath]x^2+(y+v)^2=r^2[/imath]

This yields a solution [imath]x=\dfrac{-u+\sqrt{u^2-4\alpha}}{2}, y=\sqrt{r^2-(x+u)^2}[/imath]
where [imath]\alpha=\dfrac{u^2+v^2}{4}-\dfrac{v^2r^2}{u^2+v^2}[/imath]

(Depending on whether u and/or v are less than r, there may be a corresponding solution in each of the 4 quadrants, in 2 quadrants, in only 1 quadrant or no solution if u, v are 'too large' relative to r).

Thanks @lex, for the detailed explanation. This helps a lot.

 

[shan86]

If you need to find the angles rather than coordinates, I would do this:

View attachment 28858

Using the right triangle with legs x and y, find the diagonal, and also find the angle marked in red. This is congruent to the red angle at the origin, because the sides of the angles are perpendicular.

Then using the right triangle with the red line as a leg, find the angle marked in blue. From those two angles, you can find a and b.

Thanks @Dr. Peterson, Your solution is the easiest. I think I can go with this. I'll update the result. Thanks again.

 

[shan86]

If anyone interested, this is how I implemented it in c/c++. Thanks everyone.

double l = ::sqrt(x*x + y*y) / 2.0;
double a1 = ::atan2(x, y);
double a2 = ::asin(l / r);
double a3 = a1 - a2;

_ASSERT(a1 > a2);

double b = a2 * 2;
double a = a1 - a2;