IIT JEE , AIEEE Forum - Mathematics , Trignometry

joyfrancis

Ok I'm having a real tough time solving the problems of these chapter, I've got lots of doubts which i am writing here please solve as much as you can.

1) if in a triangle ABC a⁴+b⁴+c⁴ = 2c²(a²+b²) prove that angle C is either 45 or 135

2) If I_nis the area of an n sided polygon inscribed in a circle of unit radius and O_nbe the area of the polygon circumscribing the circle prove that

I_n= O_n/2 [1+ sqroot(1-(2I_n/n)²)]

3) if the tangents of the angles of a triangle are in AP prove that the squares of the sides are in the ratio x²(x²+9) : (3+x²)²: 9(1+x²); where x is the tangent of the least or the greatest angle

4) prove delta <= 1/4 (sqroot(a+b+c)abc)

5) prove (acos²A/2 + bcos²B/2 + ccos²C/2) / (a+b+c) <= 3/4

6) If a²,b²,c² are in AP , then cotA,cotB,cotC are in ?

7) IN AN ACUTE ANGLED TRIANGLE THE LEAST VALUE OF SECA+SECB+SECC IS?

please help!

ankurgupta91

1) 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.
hope u got it nw
thanks................

ankurgupta91

6) a^2,b^2,c^2 are in A.P
so , 2b^2= a^2+c^2 ---------(1)
nw cot A = b^2+c^2-a^2/4delta
cotA = 3b^2-2a^2/delta from(1)
cot B = a^2+c^2-b^2/delta
cotB =b^2/delta
cot C= a^2+b^2-c^2/delta
cotC= 2a^2 - b^2/delta
nw from these values we get
cot B - cot A = cotC- cotB
so Cot A ,cotB ,Cot C are in A.P
thanks..........

joyfrancis

brilliant dude. Try the others also plz

ankurgupta91

7)we knw that cos A + COS B+CosC>=3/2
use A.M >= G.M
(cos A +CosB+cosC)/3 >=(cosA.cosB.cosC)^1/3
so, cosA.cosB.cosC<= 1/8
again use A.M >= G.M
(secA+secB+secC)/3>= (secA.secB.secC)^1/3
(secA+secB+secC)/3>=(1/ cosA.cosB.cosC)^1/3
(secA+secB+secC)/3>=(8)^1/3
(secA+secB+secC)>=6
thus , the least value of(secA+secB+secC) is 6
thats the rght answer
thanks...........

bhargavi

4 sol)

²=s(s-a)(s-b)(s-c)

²/s=(s-a)(s-b)(s-c)=

(s-a)(s-b)*

(s-b)(s-c)*

(s-c)(s-a).......(1)

using A.M>=G.M,

(S-a)+(s-b)/2>=

(s-a)(s-b)

{[(b+c-a)/2]+[(a+c-b)/2]}/2>=

(s-a)(s-b)

c/2>=

(s-a)(s-b)

(s-a)(s-b)<=c/2.......(2)

similarly,

(s-b)(s-c)<=a/2.......(3)

(s-c)(s-a)<=b/2........(4)

using the results (2), (3) &(4) in (1)

²/s<=(c/2)(a/2)(b/2)

²<=s(abc/8)<=[(a+b+c)/2](abc/8)=(a+b+c)(abc)/16

hence,

<=(1/4)

(a+b+c)(abc)

bhargavi

considering the polygon which is inscribed in the circle,the angle subtended by each side of the polygon at the centre of circle of unit radius is ( 2

/n). so area of each triangle is (1/2)(1)(1)sin(2

/n).

hence area of n such triangles is=(n/2)sin(2

/n)

i.e. I_n=(n/2)sin(2

/n)...........(1)

when the polygon is circumscribed, area of each such triangle=

(1/2)(1)(2tan[

/n])=tan(

/n) i.e. o_n=ntan(

/n).........(2). now from (1) &(2)

R.H.S=(n/2)tan(

/n)[1+{1-sin²(2

/n)}^1/2]=(n/2)tan(

/n)[1+cos(2

/n)]

=(n/2)sin(

/n)[2cos²(

/n)]/cos(

/n)

=(n/2)sin(2

/n)=I_n{from (1)}

hence, I_n=O_n/2[1+(1-{2I_n/n}²)^1/2]

joyfrancis

the others!!????

aysh

7)IT"S QUITE SIMPLE...
secA + secB + secC needs 2 be minimum.
i.e.,cosA, cosB, cosC need 2 be max. simultaneously.

Also, A,B,C r acute.

THUS,
minimum value of G.E.= 6

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