# Help With a Diagram

• Jun 2nd 2012, 08:56 PM
MarkGH
Help With a Diagram
Can someone help me solve the following diagram, if at all possible...

http://i1163.photobucket.com/albums/...1/Triangle.png

Solve for Angle: Theta

Regards,
- Mark
• Jun 2nd 2012, 09:58 PM
Prove It
Re: Help With a Diagram
If we call the hypotenuse of the smaller triangle \displaystyle \displaystyle \begin{align*} x \end{align*}, then our large triangle has dimensions \displaystyle \displaystyle \begin{align*} A, B + x, \sqrt{A^2 + (B + x)^2} \end{align*} (by Pythagoras).

Since the triangles are similar, their corresponding sides are in the same proportion, so

\displaystyle \displaystyle \begin{align*} \frac{A}{R} &= \frac{\sqrt{A^2 + (B + x)^2}}{x} \\ \frac{A^2}{R^2} &= \frac{A^2 + (B + x)^2}{x^2} \\ \frac{A^2}{R^2} &= \frac{A^2 + B^2 + 2Bx + x^2}{x^2} \\ A^2x^2 &= R^2\left(A^2 + B^2 + 2Bx + x^2\right) \\ A^2x^2 &= A^2R^2 + B^2R^2 + 2BR^2x + R^2x^2 \\ \left(A^2 - R^2\right)x^2 - 2BR^2x &= A^2R^2 + B^2R^2 \\ x^2 - \frac{2BR^2}{A^2 - R^2}x &= \frac{A^2R^2 + B^2R^2}{A^2 - R^2} \\ x^2 - \frac{2BR^2}{A^2 - R^2}x + \left(-\frac{BR^2}{A^2 - R^2}\right)^2 &= \frac{A^2R^2 + B^2R^2}{A^2 - R^2} + \left(-\frac{BR^2}{A^2 - R^2}\right)^2 \\ \left(x - \frac{BR^2}{A^2 - R^2}\right)^2 &= \frac{\left(A^2R^2 + B^2R^2\right)\left(A^2 - R^2\right) + B^2R^4}{\left(A^2 - R^2\right)^2} \\ \left(x - \frac{BR^2}{A^2 - R^2}\right)^2 &= \frac{A^4R^2 - A^2R^4 + A^2B^2R^2 - B^2R^4 + B^2R^4}{\left(A^2 - R^2\right)^2} \\ \left(x - \frac{BR^2}{A^2 - R^2}\right)^2 &= \frac{A^4R^2 - A^2R^4 + A^2B^2R^2}{\left(A^2 - R^2\right)^2}\end{align*}

\displaystyle \displaystyle \begin{align*} x - \frac{BR^2}{A^2 - R^2} &= \pm \frac{\sqrt{A^4R^2 - A^2R^4 + A^2B^2R^2}}{A^2 - R^2} \\ x &= \frac{BR^2 \pm \sqrt{A^4R^2 - A^2R^4 + A^2B^2R^2}}{A^2 - R^2} \end{align*}

Therefore, the height of the larger triangle is \displaystyle \displaystyle \begin{align*} B + \frac{BR^2 \pm \sqrt{A^4R^2 - A^2R^4 + A^2B^2R^2}}{A^2 - R^2} \end{align*}, and so we have the relationship

\displaystyle \displaystyle \begin{align*} \tan{\theta} &= \frac{B + \frac{BR^2 \pm \sqrt{A^4R^2 - A^2R^4 + A^2B^2R^2}}{A^2 - R^2}}{A} \\ \theta &= \tan^{-1}{\left(\frac{B + \frac{BR^2 \pm \sqrt{A^4R^2 - A^2R^4 + A^2B^2R^2}}{A^2 - R^2}}{A}\right)} \end{align*}
• Jun 3rd 2012, 02:31 AM
BobP
Re: Help With a Diagram
Here's an alternative method.

First some construction. (Apologies, I don't know how to incorporate diagrams, maybe one of the sites resident experts could suggest a method ?)

Run a line from the centre of the circle parallel with the hypotenuse down to the base of the triangle. Call the three points along the base D, E and F. (D is at the end of the hypotenuse, E is the new point and F is at the rightangle). Run another line from E upto the hypotenuse to meet it at rightangles (it will be parallel with the line in the circle), and call this point G.

In the new triangle DEG, $\displaystyle GE/DE=\sin\theta$, but $\displaystyle GE=R,$ so $\displaystyle DE=R/\sin\theta,$

in which case$\displaystyle EF=A-R/\sin\theta.$

Therefore, $\displaystyle \frac{B}{A-\frac{R}{\sin\theta}}=\tan\theta=\frac{sin\theta}{ \cos\theta}.$

Mutiply top and bottom of the LHS by $\displaystyle \sin\theta$ to get $\displaystyle \frac{B\sin\theta}{A\sin\theta-R}=\frac{\sin\theta}{\cos\theta},$

cancel the $\displaystyle \sin\theta$ 's, cross multiply, rearrange, and you arrive at $\displaystyle A\sin\theta-B\cos\theta=R.$

From hereon it's routine, put the LHS equal to $\displaystyle K\sin(\theta-\alpha),$ expand and equate with the existing LHS to get

$\displaystyle K\cos\alpha = A,\quad K\sin\alpha=B$ so that $\displaystyle K=\sqrt{A^{2}+B^{2}}$ and $\displaystyle \tan\alpha=B/A$.

Finally then, $\displaystyle R=\sqrt{A^{2}+B^{2}}\sin(\theta-\alpha),$

$\displaystyle \theta=\arcsin\frac{R}{\sqrt{A^{2}+B^{2}}}+\arctan \frac{B}{A}.$

S*** what an idiot !!!

Draw a line from the centre of the circle down to the base of the hypotenuse. The "new" hypotenuse will be of length $\displaystyle \sqrt{A^{2}+B^{2}}$.
The angle $\displaystyle \theta$ will be the sum of those two angles which will be, from the two triangles $\displaystyle \arcsin\frac{R}{\sqrt{A^{2}+B^{2}}}$ and $\displaystyle \arctan\frac{B}{A}.$
• Jun 14th 2012, 12:42 PM
richard1234
Re: Help With a Diagram
Draw the segment containing the two right angles. We'll call it PQ, where P is the right angle in the smaller triangle and Q is the right angle in the larger triangle. Suppose the three vertices of the large right triangle are, in clockwise order, QRS, where Q is a right angle.

The hypotenuse of the large triangle is given by $\displaystyle RS = \frac{A}{\cos \theta}$. Also, $\displaystyle PS = R \tan \theta$.

Therefore, $\displaystyle RP = RS - PS = \frac{A}{\cos \theta} - R \tan \theta = \frac{A - R \sin \theta}{\cos \theta}$

Applying the law of cosines,

$\displaystyle PQ^2 = R^2 + B^2 - 2RB \cos (180 - \theta)$

$\displaystyle PQ^2 = A^2 + (\frac{A - R \sin \theta}{\cos \theta})^2 - 2A(\frac{A - R \sin \theta}{\cos \theta}) \cos \theta$

Equate these two, noting that $\displaystyle \cos (180 - \theta) = -\cos \theta$

$\displaystyle R^2 + B^2 + 2RB \cos \theta = A^2 + (\frac{A - R \sin \theta}{\cos \theta})^2 - 2A(\frac{A - R \sin \theta}{\cos \theta}) \cos \theta$

And so on.