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In a triangle ABC, what is the angle $\theta$ between the median AM and the bisector AD? I want a way to know the measure of that angle, given the lenghts of the sides and angles of the triangle. I tried solving the triangle, with the sine and cosine laws, and I arrived at $$ \theta = \frac{A}{2} - \arcsin\left(\frac{2\sqrt{(a/2)^2+b^2 -ab\cos C}} {a\sin C}\right). $$ I'm not even totally sure if it's correct and it's a bit unwieldy... I am wondering if there is a smarter aproach that gives a nicer formula. Thank you!

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  • $\begingroup$ What is $\Gamma?$ $\endgroup$ – saulspatz Apr 29 '18 at 17:12
  • $\begingroup$ sorry! I mean angle C, I wasn't consistent in the notation $\endgroup$ – George Ntoulios Apr 29 '18 at 17:14
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    $\begingroup$ See this. $\endgroup$ – MalayTheDynamo Apr 29 '18 at 17:24
  • $\begingroup$ nice! with this the sqare root in the formula becomes $\sqrt{\frac{b^2+c^2}{2}-a^2}$ it is much better for sure $\endgroup$ – George Ntoulios Apr 29 '18 at 17:41
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Let $A = 2\alpha$, $B = 2\beta$, $C = 2\gamma$. We can coordinatize, with $$A = (0,0) \qquad B = c\left(\cos\alpha,-\sin\alpha\right) \qquad C = b\left(\cos\alpha,\sin\alpha\right) \qquad M = \frac12(B+C)$$

Since bisector $\overline{AD}$ coincides with the $x$-axis, $\angle DAM$ is simply the direction angle of $M$. Perhaps the simplest representation is

$$\tan DAM = \frac{M_y}{M_x} = \frac{B_y+C_y}{B_x+C_x} = \frac{(b-c)\sin\alpha}{(b+c)\cos\alpha} \tag{$\star$}$$

The Law of Sines tells us that $b = d \sin B$ and $c = d \sin C$, where $d$ is the circumdiameter of $\triangle ABC$. With trig's prosthaphaeresis formulas, and recalling that $\alpha + \beta + \gamma = 90^\circ$, we can rewrite $(\star)$ as

$$\tan DAM = \frac{\sin^2\alpha \sin(\beta-\gamma)}{\cos^2\alpha\cos(\beta-\gamma)} = \tan^2\alpha \tan(\beta-\gamma) \tag{$\star\star$}$$

Alternatively, using the half-angle formulas, we have $$\tan DAM = \frac{(1-\cos A)\sqrt{1-\cos(B-C)}}{(1+\cos A)\sqrt{1+\cos(B-C)}}$$ which, with the help of the Law of Cosines, is manipulatable into this lengths-only form

$$\tan DAM = \frac{b-c}{b+c}\;\sqrt{\frac{\phantom{-}(a - b + c)(a + b - c)}{(-a + b + c) (a + b + c)}} \tag{$\star\star\star$}$$

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By law of sines: $$ \frac{\sin CAM}{\sin BAM}=\frac{\sin C}{\sin B}=:\lambda. $$

Setting $\sin BAM=x$, and solving the resulting equation for cosine of the sum: $$ \sqrt{1-x^2}\cdot\sqrt{1-\lambda^2x^2}-x\cdot\lambda x=\cos A\\ $$ one obtains: $$ \sin BAM=\frac{\sin A}{\sqrt{1+\lambda^2+2\lambda\cos A}} =\frac{\sin A\sin B}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}} $$ and similarly $$ \sin CAM=\frac{\sin A\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}. $$ Correspondingly: $$ \cos BAM=\frac{\sin B\cos A+\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}},\\ \cos CAM=\frac{\sin B+\sin C\cos A}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}. $$

Writing: $$\small{ \cos(BAM-CAM)=\frac{(\sin B\cos A+\sin C)(\sin B+\sin C\cos A)+\sin B\sin C\sin^2A}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}\\ =\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A} } $$ and noticing that $|\widehat{BAM}-\widehat{CAM}|$ is the doubled value of the angle in question one finally obtains: $$ \theta=\frac{1}{2}\arccos\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}. $$

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