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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,400

1. If the sides of a triangle are in Arithmetic Progression, then find the value of

in terms of

.

2. A circle is inscribed in an equilateral triangle of side a. Find the area of the square inscribed in this circle.

3. Find the general value of θ satisfying the equation

tan[sup]2[/sup]θ + sec2θ = 1.

4. Solve the equation

sin x - 3sin 2x + sin 3x = cos x - 3cos 2x + cos 3x.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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5. Find the area bounded by the curve x[sup]2[/sup]=4y and the

straight line x = 4y - 2.

6. If a, b, c are the three sides of a triangle and C = 60°, prove that

7. If a>0, b>0, and c>0, prove that

8.Find the equations of straight lines passing through (-2, -7) and having an interept of length 3 between the striaght lines

4x+3y = 12 and 4x+3y = 3.

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**JaneFairfax****Member**- Registered: 2007-02-23
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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,400

Answer to 7:-

Correct, JaneFairfax!

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,400

9. For what values of m does the system of equations

3x + my = m and

2x - 5y = 20

have solutions satisfying the condtions x>0, y>0?

10. Let -1 ≤ p ≤ 1. Show that the equation 4x[sup]3[/sup] - 3x - p = 0 has a unique root in the interval [½, 1] and identify it.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Ganesh;

For #9

Solve for x and y in terms of m to understand how x and y behave when m varies:

After a lot of algebraic thrashing and some trial and error:

These values of m satisfy the constraints x>0 and y>0

*Last edited by bobbym (2009-04-13 06:38:13)*

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

#1: tan(A/2) +tan(C/2) = (2/3)cot(B/2)

#4: x=pi/8 or 3pi/8

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

#6 : cosC=1/2⇒b² -(b/2)²=c² -(a-b/2)² ⇒ab=a²+b²-c² ⇒3ab=(a+b)² -c²=(a+b+c)(a+b-c)=(a+b+c)(a+b+c-2c)=

(a+b+c)²-2c(a+b+c)⇒3ab+2c(a+b+c)=(a+b+c)²⇒3ab+3c(a+b+c)=(a+b+c)²+c(a+b+c)⇒

3(a+c)(b+c)=(a+b+c)(b+c+a+c)⇒[(b+c)+(a+c)]/(a+c)(b+c)=3/(a+b+c)⇒1/(a+c)+1/(b+c)=

3/(a+b+c).

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi bitus,

For #4 I think you mean

doesn't work when I plug it in.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**BO****Member**- Registered: 2010-05-16
- Posts: 9

1.from sine rule we get sinA+sinC=2sinB, or cos([A-C]/2)=2cos(B/2)

tan(A/2)+tan(c/2)=cos(B/2)/[cos (A/2) cos (C/2)]=2cos(B/2)/[sin(B/2)+2cos(B/2)]=2/[1+cot (B/2)]

2.the radius of the circle=a*tan30=a/3

Then the diameter of the square=2a/3

area=(2a/3 sin45)^2=a^2/9

*Last edited by BO (2010-05-16 19:32:48)*

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**BO****Member**- Registered: 2010-05-16
- Posts: 9

3. tan θ=1/2

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**irspow****Member**- Registered: 2005-11-24
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*Last edited by irspow (2010-05-21 03:04:34)*

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,400

Well done,

irspow!!!

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