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Determine the ratio of two angles in a triangle.

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Let \(\displaystyle ABC\) be a triangle such that \(\displaystyle \frac{BC}{AB-BC}=\frac{AB+BC}{AC}\).

Determine the ratio \(\displaystyle \frac{\angle A}{\angle C}\).
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let \(\displaystyle ABC\) be a triangle such that \(\displaystyle \frac{BC}{AB-BC}=\frac{AB+BC}{AC}\).

Determine the ratio \(\displaystyle \frac{\angle A}{\angle C}\).
Write $a,\ b,\ c$ for the sides opposite the angles $A,\ B,\ C$ respectively. You are told that $\dfrac a{c-a} = \dfrac{c+a}b$, so that $ab = c^2-a^2$. The cosine rule tells you that $c^2 = a^2+b^2-2ab\cos C$. Combining those equations, you get $ab = b^2 -2ab\cos C$, so that $a = b-2a\cos C$. Now draw a picture.


The base of the isosceles triangle $BCD$ is twice the projection of $BC$ onto the side $CA$, namely $2a\cos C$. Therefore $DA = b-2a\cos C = a$. Thus $BDA$ is also isosceles, and it follows that $\angle C = \angle BDC = 2\angle A$.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Write $a,\ b,\ c$ for the sides opposite the angles $A,\ B,\ C$ respectively. You are told that $\dfrac a{c-a} = \dfrac{c+a}b$, so that $ab = c^2-a^2$. The cosine rule tells you that $c^2 = a^2+b^2-2ab\cos C$. Combining those equations, you get $ab = b^2 -2ab\cos C$, so that $a = b-2a\cos C$. Now draw a picture.


The base of the isosceles triangle $BCD$ is twice the projection of $BC$ onto the side $CA$, namely $2a\cos C$. Therefore $DA = b-2a\cos C = a$. Thus $BDA$ is also isosceles, and it follows that $\angle C = \angle BDC = 2\angle A$.
Awesome! :cool: I used a totally different method and yours is without a doubt, so much better than mine!

Thanks for showing me that we could actually solve a geometric problem using such a method and to be honest with you, I always love to read how you would approach a problem...:eek:

Thank you so much, Opalg!