Question

ML - Question of the DAY : [1] - 20 / 04 / 2013

With usual notation, if in a triangle ABC : \(\cfrac{(b+c)}{11} = \cfrac{c+a}{12} = \cfrac{a+b}{13} \) . Then prove that :

\(\cfrac{\cos A}{7} = \cfrac{\cos B}{19} = \cfrac{\cos C}{25} \)

Answer 1

NOTE : There will be no hints , though in case of harder questions, I will provide hints.

Please provide your answer here either with complete solution or with some of your steps .

BEST OF LUCK! [today's one is easy]

Answer 2

$$\cfrac{(b+c)}{11} = \cfrac{c+a}{12} = \cfrac{a+b}{13}=t$$
$$b+c=11t\\
c+a=12t\\
a+b=13t $$
from these three equations,
$$b=6t,\quad c=5t,\quad a=7t$$
$$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{36+25-49}{2\times6\times5}=1/5$$
similarly,
$$\cos B=19/35,\quad\cos C=5/7$$
$$\frac{\cos A}{\cos B}=\frac{1/5}{19/35}=\frac{7}{19}\implies \frac{\cos A}{7}=\frac{\cos B}{19}$$
$$\frac{\cos A}{\cos C}=\frac{1/5}{5/7}=\frac{7}{25}\implies \frac{\cos A}{7}=\frac{\cos C}{25}$$
$$\therefore \frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$$

Answer 3

Excellent work @BAdhi .

\(\textbf{ML - QUESTION OF THE DAY - 20/04/2013 - SOLVED BY BADHI}\)

\(\checkmark\) \(\checkmark\) \(\checkmark\) \(\checkmark\) \(\checkmark\)\(\checkmark\)

Answer 4

This question will be closed on 21-04-2013 .

Thanks!

Answer 5

Thanks for the complements. Glad to work out a quest which is quite challenging. hoping to see more in future :)

Answer 6

Glad to hear , that you liked it. Yes BAdhi, I will try to post these questions every day.

Answer 7

a,b,c stand for sides, A,B,C stand for angles?

Answer 8

Yes perl.