Circle with a triangle poking out of it.

[Valentine988]

I'm not even sure if my math is right:
Sin45°= r / (r+15)

r = (r+15)(Sin45°)

r = rSin45° + 15Sin45°

Any help is greatly appreciated.

 

[pka]

Do you know that $\displaystyle \Delta ACD$ is a right triangle and because $\displaystyle \angle A = 45$ $\displaystyle \Delta ACD$ is an isosceles triangle?

 

[jonboy]

Do you know that $\displaystyle \Delta ACD$ is a right triangle and because $\displaystyle \angle A = 45$ $\displaystyle \Delta ACD$ is an isosceles triangle?

Very true. So you could do Law of Sines:\(\displaystyle \L \;\frac{Sin90}{15\,+\,r}\,=\,\frac{Sin45}{r}\)

\(\displaystyle \L \;\frac{1}{15}\,=\,\frac{.7071}{r}\)

Solve for $\displaystyle r$

Or with Valentine988's method find the value for Sin and solve for $\displaystyle r$

Find value by inserting the degree angle and pressing the Sin button.

 

[pka]

Or just note that $\displaystyle \left| {AD} \right| = r \Rightarrow \quad r^2 + r^2 = \left( {x + r} \right)^2 .$

 

[Denis]

Jonboy, shouldn't that be SIN(90) / (15 + r)?
And then, why use SIN?

 

[jonboy]

Jonboy, shouldn't that be SIN(90) / (15 + r)?
And then, why use SIN?

Yes, I used Sin because it works.

 

[Shriya]

In the given triangle CDA,
A=45
sin 45=CD/CA
=r/r+x
=r/r+15
sin45=0.707(approx)=r/r+15
therefore, 0.707(r+15)=r

r=36.24(approx)

 

[Denis]

Well, whatever...

You've got a right triangle, hypotenuse = r+15, both legs = r;
perfect candidate for Pythagoras

 

[jonboy]

Well, whatever...

You've got a right triangle, hypotenuse = r+15, both legs = r;
perfect candidate for Pythagoras

Yeah that's good.