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Trignometry

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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 m

C is equal to

(a) 60⁰ (b) 135⁰

Comments (3)

ankur gupta

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Joined: 10 Mar 2007

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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................

Sanchay Gupta

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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

Chirag...I rock.. ..

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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

adding 2*a^2*b^2 both sides,

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)

(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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