Given acute Triangle $A,B,C$ and angle $\theta$ opposite $b$ and angles $ \alpha$ and $ \beta$ where $\alpha +\beta = \theta$ created by the perpendicular from $B$ to the opposite side,

Find: $\frac{\cos(\alpha)}{\sin(\theta)} $enter image description here

Finding this is not too bad the work is shown below.
My question is what happens if it is not a perpendicular line. Suppose the ratio of the divided sides was $k$. Is there a simple form solution?

Solution: Let $h$ be the length of the height.

Then: $cos(\alpha) = \frac{h}{c}$ and $cos(\beta) = \frac{h}{a}$. Therefore $$\frac{\cos(\alpha)}{\cos(\beta)} = \frac{c}{a} \Rightarrow \cos(\beta) = \frac{a}{c} \cos(\alpha) $$ $$\cos(\theta) = \cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$$

Let $\frac{a}{c} = K$

$$\cos(\theta) = K\cos(\alpha)^{2} - \sin(\alpha)\sin(\beta)$$ $$cos(\theta) - K\cos(\alpha)^{2} = \sqrt{1-cos(\alpha)^2}\sqrt{1-cos(\beta)^2}$$ $$\left( cos(\theta) - K\cos(\alpha)^{2} \right)^2 = \left( 1-cos(\alpha)^2 \right)\left(1 -K^2 cos(\alpha)^2\right)$$

$$\cos(\theta)^2 -2K\cos(\alpha)^2\cos(\theta)+K^2\cos(\alpha) = 1 -(1+K^2)\cos(\alpha)^2 + K^2\cos(\alpha) $$ $$-\sin(\theta)^2 = 2K\cos(\alpha)^2\cos(\theta) - \left ( {1+K^{2} }\right)\cos(\alpha)^2 $$

$$\frac{-sin(\theta)^2}{2Kcos(\theta) -1 - K^2} = cos(\alpha)^2$$ $$\frac{-\frac{1}{k} sin(\theta)^2}{2cos(\theta) -\frac{1}{K} - K} = -\frac{\frac{c}{a} sin(\theta)^2}{2cos(\theta) -\frac{c}{a} -\frac{a}{c} } = \frac{-c^2 sin(\theta)^2}{(2ca)\cos(\theta) -a^2 -c^2}=cos(\alpha)^2$$

$$\frac{c^2 \sin(\theta)^2}{b^2} = \cos(\alpha)^2$$ $$\cos(\alpha) = \frac{c\sin(\theta)}{b}$$

By similar methods we get: $$ \cos(\beta) = \frac{a\sin(\theta)}{b}$$

$$\frac{\cos(\alpha)}{\sin(\theta)} = \frac{c}{b} \hspace{1cm} \frac{\cos(\beta)}{\sin(\theta)} = \frac{a}{b} $$


Introduce the second altitude as shown.

enter image description here

Use the areas of the same triangle but with different altitudes to calculate the ratio. (Or doing the above in the reverse order.) The proofing process will then be much simpler.

For example, red $= c \cos \theta$ and area $= \dfrac {red \cdot a}{2}$.


Use Law of Sines

$cos(\alpha)$ = sine of angle opposite to side $a$ = $sinA$

So, $sinA/sinB = a/b$ $

$(\theta=B) $

Proof is here for law of sines.


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