# the side of a triangle ABC are in A.P.(order being a,b,c) and satisfy 2!/(1!9!) + 2!/(3!7!)+ 1/(5!5!) = 8^a/(2b)! then the value of cosA+cosB is ?? options area. 12/7b. 13/7c. 11/7d. 10/7

siddharth gupta
28 Points
9 years ago
10C1=10!/1!9! i.e 1/1!9!=10C1/10!i.e2!/1!9!=210C1/10!
similarly 2!/3!7!=210C3/(10)!
and 1/5!5!=10C5/10!
the given equation can be rewritten as
(1/10!)(210C1+10C5+210C3)=(1/(2b)!)(8a)
=(1/10!)(512)=(1/(2(5)!)(83)
on comparison we can say that a=3 and b=5 and using the concept of A.P c=7.
use cosine rule to find cosA and cosB
cosA=13/14 AND cosB=11/14
COS(A)+COS(B)=24/14=12/7
WITH REGARDS
SIDDHARTH GUPTA
S
Sumit Majumdar IIT Delhi
9 years ago
Dear student,
Since the sides are in A.P. we have:
2b=a+c ...(i)
Also we know that:
$_{r}^{n}\textrm{C}=\frac{n!}{r!\left ( n-r \right )!}$
Using this, we get:
$_{1}^{10}\textrm{C}=\frac{10!}{1!9!}$
This gives,
$1!9! =\frac{10!}{_{1}^{10}\textrm{C}}$
Similarly,
$3!7! =\frac{10!}{_{3}^{10}\textrm{C}}$
and,
$5!5! =\frac{10!}{_{5}^{10}\textrm{C}}$
Hence the given equation can be rewritten as:
$\left ( \frac{1}{10!} \right )\left ( 2_{1}^{10}\textrm{C}+_{5}^{10}\textrm{C}+2_{3}^{10}\textrm{C} \right )=\left (\frac{1}{2b!} \right )\left ( 8^{3} \right )$
This yields:
$\left ( \frac{1}{10!} \right )\left ( 512\right )=\left (\frac{1}{2b!} \right )\left ( 8^{3} \right )$
Thus by comparison, we get:
$b=5$.
Since, the sides are in AP, so we have:
a+c=10
Regards
Sumit