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Trignometry

Hot goIITian

 Joined: 12 Mar 2007 Post: 189
29 Mar 2007 15:34:21 IST
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631
find angle plzzzzzzz
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

In a ABC , if a4 + b4 + c4 = 2c2(a2 + b2) , then  mC is equal to

(a) 600                       (b) 1350

(c)  900                      (d) 750

Scorching goIITian

Joined: 10 Mar 2007
Posts: 257
29 Mar 2007 15:42:25 IST
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cosine rule: c^2 = a^2 + b^2 -2abcosC
a^4 + b^4 = 2c^2(a^2+b^2 ) -c^4
a^4 + b^4 = c^2 ( 2a^ +2b^2 -c^2)
putting the value of c^2
a^4+b^4 = (a^2 + b^2 -2abcosC)(a^2+b^2+2abcosC)
a^4 + b^4=(a^2+ b^2)^2 - (2abcosC)^2
2a^2b^2 = 4a^2b^2cos^2C
cos C = +- 1/sqroot of 2
thus, angle C is either 45 or 135.
so the ans is (B)
hope u got it nw
thanks................

Hot goIITian

Joined: 12 Mar 2007
Posts: 189
30 Mar 2007 22:05:07 IST
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thanks ankur ur answer is absolutely correct

can i ask u a question
r u student of class 11 or 12 or doing college
reply if u have time otherwise its no special

Blazing goIITian

Joined: 10 Feb 2007
Posts: 517
2 Apr 2007 13:38:33 IST
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a^4+b^4+c^4=2c^2(a^2+b^2)

a^4+b^4+c^4-2c^2(a^2+b^2)=0

a^4+b^4+c^4-2c^2(a^2+b^2)+2*a^2*b^2=2*a^2*b^2

(a^2+b^2-c^2)^2=2*a^2*b^2

taking sqrt both sides,

(a^2+b^2-c^2)=+2*a*b.............(i)

or

(a^2+b^2-c^2)=-2*a*b............(ii)

using cosine rule,

cos(c)= b^2+a^2-c^2/2ab...................(iii)

Substituting (i) and (ii)

we get,

cos c= 45 or 135

therefore ans(B)

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