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$\textrm{The problem is as follows:}$

$\textrm{Find the value of H, in triangle ABC}$,

$\overline{BC}=3\overline{AC}$

Schematic of the problem

$$H=\frac{\sin2\omega\tan(\omega-\phi)}{\tan(\omega+\phi)\sin2\phi}$$

What I did was to "decompose" the above statement into the following formulas,

$$H=\frac{(2\sin\omega\cos\omega)(\frac{\tan\omega-\tan\phi}{1+\tan\omega\tan\phi})}{(\frac{\tan\omega+\tan\phi}{1-\tan\omega\tan\phi})(2\sin\phi\cos\phi)}$$

therefore,

$$H=\frac{2\sin\omega\cos\omega(\tan\omega-\tan\phi)(1-\tan\omega\tan\phi)}{(1+\tan\omega\tan\phi)(\tan\omega+\tan\phi)2\sin\phi\cos\phi}$$

Simplifying terms and replacing with $\frac{\sin\omega}{\cos\omega}$ and $\frac{\sin\phi}{\cos\phi}$ in the above equation I got to this:

$$H=\frac{(\sin\omega\cos\omega)(\frac{\sin\omega}{\cos\omega}-\frac{\sin\phi}{\cos\phi})(1-\frac{\sin\omega}{\cos\omega}\frac{\sin\phi}{\cos\phi})}{(\frac{\sin\omega}{\cos\omega}+\frac{\sin\phi}{\cos\phi})(1+\frac{\sin\omega}{\cos\omega}\frac{\sin\phi}{\cos\phi})(\sin\phi\cos\phi)}$$

$$H=\frac{(\sin\omega\cos\omega)(\frac{\sin\omega\cos\phi-\sin\phi\cos\omega}{\cos\omega\cos\phi})(\frac{\cos\omega\cos\phi-\sin\omega\sin\phi}{\cos\omega\cos\phi})}{(\frac{\sin\omega\cos\phi+\sin\phi\cos\omega}{\cos\omega\cos\phi})(\frac{\cos\omega\cos\phi+\sin\omega\sin\phi}{\cos\omega\cos\phi})(\sin\phi\cos\phi)}$$

By multiplying $\cos^{2}\omega\cos^{2}\phi$ in numerator and denominator I reached to this:

$$H=\frac{(\sin\omega\cos\omega)(\frac{\sin\omega\cos\phi-\sin\phi\cos\omega}{\cos\omega\cos\phi})(\frac{\cos\omega\cos\phi-\sin\omega\sin\phi}{\cos\omega\cos\phi})}{(\frac{\sin\omega\cos\phi+\sin\phi\cos\omega}{\cos\omega\cos\phi})(\frac{\cos\omega\cos\phi+\sin\omega\sin\phi}{\cos\omega\cos\phi})(\sin\phi\cos\phi)}\times\frac{\cos^{2}\omega\cos^{2}\phi}{\cos^{2}\omega\cos^{2}\phi}$$

therefore,

$$H=\frac{(\sin\omega\cos\omega)(\sin\omega\cos\phi-\sin\phi\cos\omega)(\cos\omega\cos\phi-\sin\omega\sin\phi)}{(\sin\omega\cos\phi+\sin\phi\cos\omega)(\cos\omega\cos\phi+\sin\omega\sin\phi)(\sin\phi\cos\phi)}$$

From the attached figure I got by using sines law this equation:

$$\frac{3k}{\sin2\omega}=\frac{k}{\sin2\phi}$$

Being $\textrm{k}$ from:

$$\overline{BC}=3\overline{AC}$$

therefore,

$$\frac{\overline{BC}}{\overline{AC}}=\frac{3\times k}{k}$$

$$\overline{BC}=3k$$

$$\overline{AC}=k$$

as a result by inserting the above equation into $\textrm{H}$ I got to this expression:

$$\frac{2\sin\omega\cos\omega}{2\sin\phi\cos\phi}=\frac{3k}{k}$$

$$\frac{\sin\omega\cos\omega}{\sin\phi\cos\phi}=3$$

and,

$$H=\frac{(3)(\sin\omega\cos\phi-\sin\phi\cos\omega)(\cos\omega\cos\phi-\sin\omega\sin\phi)}{(\sin\omega\cos\phi+\sin\phi\cos\omega)(\cos\omega\cos\phi+\sin\omega\sin\phi)}$$

However I'm stuck at here. I don't see other expression cancelling. What should be done to solve this problem?.

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  • $\begingroup$ @Rohan Mind explaining how did you got to that expression?. Moreover How does it help to solve the problem?. I don't understand that part. $\endgroup$ – Chris Steinbeck Bell Dec 9 '17 at 10:23
  • $\begingroup$ @Rohan. Sorry I had a mistake in re-drawing the figure from the book, the image is incorrect, it should be C instead of B. I'll be re uploading the figure to correct that situation. $\endgroup$ – Chris Steinbeck Bell Dec 9 '17 at 10:45
  • $\begingroup$ @Rohan The picture has been corrected, if you spot any errors please write them in comments. But does this affects to your earlier formula? $\endgroup$ – Chris Steinbeck Bell Dec 9 '17 at 11:03
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$$H=\frac{\sin2\omega\tan(\omega-\phi)}{\tan(\omega+\phi)\sin2\phi}$$

Note that $\frac{\sin(2\phi)}{|AC|} = \frac{\sin(2\omega)}{|BC|} \Leftrightarrow 3=\frac{\sin(2\omega)}{\sin(2\phi)} $

$$ H=3 \cdot \frac{\tan(\omega-\phi)}{\tan(\omega+\phi)} =3\cdot \frac{\sin(2\omega)-\sin(2\phi)}{\sin(2\omega) +\sin(2\phi)}=\boxed{\frac{3}{2}}$$

The simplification for $ \frac{\tan(\omega-\phi)}{\tan(\omega+\phi)}=\frac{\sin(2\omega)-\sin(2\phi)}{\sin(2\omega) +\sin(2\phi)}$ is left as an excercise :)

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  • $\begingroup$ It took me a while to reach the simplification you had left as an exercise but it looks that I was on the right track from my steps as noted in the question. All i had to do was to expand $H$ and then associate similar terms and simplify $\frac{1}{2}$, finally execute the multiplication by $3$, as such I'm adding the procedure as another answer. Thanks by the way to notice that. $\endgroup$ – Chris Steinbeck Bell Dec 9 '17 at 14:59
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I am adding this as a suggested answer from the follow up to the question I posted.

$$H=3\times \frac{(\sin\omega\cos\phi-\sin\phi\cos\omega)(\cos\omega\cos\phi-\sin\omega\sin\phi)}{(\sin\omega\cos\phi+\sin\phi\cos\omega)(\cos\omega\cos\phi+\sin\omega\sin\phi)}$$

$\textrm{in numerator:}$

$$\sin\omega\cos\omega\cos^{2}\phi-\cos\phi\sin\phi\sin^{2}\omega-\sin\phi\cos\phi\cos^{2}\omega+\sin^{2}\phi\cos\omega\sin\omega$$

$$(\cos^{2}\phi+\sin^{2}\phi)(\sin\omega\cos\omega)-(\sin^{2}\omega+\cos^{2}\omega)(\cos\phi\sin\phi)$$

$$(1)(\sin\omega\cos\omega)-(1)(\cos\phi\sin\phi)$$

$\textrm{in denominator:}$

$$(\sin\omega\cos\phi+\sin\phi\cos\omega)(\cos\omega\cos\phi+\sin\omega\sin\phi)$$

$$\sin\omega\cos\omega\cos^{2}\phi+\sin\phi\cos\phi\sin^{2}\omega+\sin\phi\cos\phi\cos^{2}\omega+\cos\omega\sin\omega\sin^{2}\phi$$

$$(\cos^{2}\phi+\sin^{2}\phi)(\sin\omega\cos\omega)+(\sin^{2}\omega+cos^{2}\omega)(\sin\phi\cos\phi)$$

$$(1)(\sin\omega\cos\omega)+(1)(\sin\phi\cos\phi)$$

therefore by rearranging both as a fraction I got to this

$$H= 3 \times \frac{(1)(\sin\omega\cos\omega)-(1)(\cos\phi\sin\phi)}{(1)(\sin\omega\cos\omega)+(1)(\sin\phi\cos\phi)}$$

$$H= 3 \times \frac{(\sin\omega\cos\omega)-(\cos\phi\sin\phi)}{(\sin\omega\cos\omega)+(\sin\phi\cos\phi)}$$

$$H= 3 \times \frac{(\frac{1}{2})(2\times\sin\omega\cos\omega)-(\frac{1}{2})(2 \times\cos\phi\sin\phi)}{(\frac{1}{2})(2 \times \sin\omega\cos\omega)+(\frac{1}{2})(2 \times \sin\phi\cos\phi)}$$

$$H= 3 \times \frac{(\frac{1}{2})(\sin2\omega)-(\frac{1}{2})(\sin2\phi)}{(\frac{1}{2})(\sin2\omega)+(\frac{1}{2})(\sin2\phi)}$$

$$H= 3 \times \frac{(\frac{1}{2})(\sin2\omega-\sin2\phi)}{(\frac{1}{2})(\sin2\omega+\sin2\phi)}$$

$$H= 3 \times \frac{(\sin2\omega-\sin2\phi)}{(\sin2\omega+\sin2\phi)}$$

Therefore by calling the relationship from sines law:

$$\sin2\omega=3\times\sin2\phi$$

$$H= 3 \times (\frac{3\sin2\phi-\sin2\phi}{3\sin2\phi+\sin2\phi})$$

$$H= 3 \times (\frac{2\sin2\phi}{4\sin2\phi})$$

$$H= 3 \times (\frac{1 \times \sin2\phi}{2 \times \sin2\phi})$$

$$H= \frac{3\sin2\phi}{2\sin2\phi}$$

Thus H becomes into:

$$H= \frac{3}{2}$$

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