Trigonometry> the side of a triangle ABC are in A.P.(or...

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 are
a. 12/7
b. 13/7
c. 11/7
d. 10/7

Rohit Falake , 11 Years ago

Grade 12th pass

2 Answers

siddharth gupta

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)!)(8^a)

=(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

Last Activity: 11 Years ago

Sumit Majumdar

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

This yields, the required answer.

Regards

Sumit

Last Activity: 11 Years ago

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Trigonometry> the side of a triangle ABC are in A.P.(or...

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

Rohit Falake , 11 Years ago

siddharth gupta

Sumit Majumdar

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