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I have two RA triangles with a shared common hypotenuse. Given the lengths of the adjacent sides of the two right angle triangles, a1 and a2, and the sum of both angles, theta, how can I calculate each angle, theta1 and theta2?

shared hypotenuse http://www.evobox.com/SharedHypotenuse.png

Obviously each angle is inversely proportionate to the length of adjacent sides, and when each side is equal the two angles are equal and half the sum of both angles. When the sides are unequal so are the angles but the relationship is non-linear. So how do I calculate each angle?

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$\cos\theta_1 = \dfrac{a_1}{c}$.

Then, \begin{eqnarray*} \cos\theta_2 &=& \frac{a_2}{c} \\ && \\ &=& \frac{a_2}{a_1}\cos\theta_1 \\ && \\ &=& \frac{a_2}{a_1}\cos\left(\theta - \theta_2\right) \\ && \\ &=& \frac{a_2}{a_1}\left(\cos\theta \cos\theta_2 + \sin\theta \sin\theta_2 \right) \\ && \\ 1 &=& \frac{a_2}{a_1}\left(\cos\theta + \sin\theta \tan\theta_2 \right) \\ && \\ \tan\theta_2 &=& \frac{\dfrac{a_1}{a_2} - \cos\theta}{\sin\theta} \\ && \\ \theta_2 &=& \tan^{-1}\left( \frac{\dfrac{a_1}{a_2} - \cos\theta}{\sin\theta} \right). \end{eqnarray*}

Now we have $\theta_2$, we have $\theta_1 = \theta - \theta_2$.

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    $\begingroup$ A minor improvement:- $θ_1 = θ - θ_2$ $\endgroup$ – Mick Jul 3 '14 at 15:00
  • $\begingroup$ @Mick Ha ha, yes, of course. Thanks. I've made the change. $\endgroup$ – Mick A Jul 3 '14 at 15:19
  • $\begingroup$ Thanks Mick for your answer, it is immensely helpful. And I've learnt something I've been trying to solve for months. $\endgroup$ – user155321 Jul 7 '14 at 9:54
  • $\begingroup$ @user155321 I'm glad I could help. $\endgroup$ – Mick A Jul 7 '14 at 13:39

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