qstion on g.p. and trigonometry

mukesh · 2010-07-28 20:32:51

ques1----- if a,b,c are pth ,qth,rth terms of an a.p. and g.p.both prove that
a^b *b^c *c^a=a^c *b^a *c^b

ques2-------from three nuimers in g.p. other three numbersin g.p are subtracted and the remainder are also found to be in g.p.prove that the three sequences have the same common ratio.

ques3-------the sides of a quadrilatral are 3,4,5,6cms.the sum of a pair of opposites angles is 120degree show that the area of quadrilatral is 16.431 sq cm. PLEASE HELP IMMEDIATILY AND THANK U IN ADVANCE.

Bob · 2010-07-28 21:39:14

Hi Mukesh

Is this your maths homework? I'll give you a hint to get you started, but, I think, you need to do most of the work yourself or you'll never improve. (Sorry, it's 43 years of being a maths teacher that's done that to me!)

Question 1. As they're in ap and gp you should be able to make two formulas connecting a, b and c by writing successive terms with 'd' and 'r' for the difference and ratio and eliminating these unknowns.

Then form the expression (a^b*b^c*c^a) / (a^c*b^a*c^b). You want to prove this = 1.

Use laws of indices eg . x^m / x^n = x^(m-n).

Simplify and use your formulas and you should be able to simplify all the way to '1' .

Question 2. Again, lots of algebra and simplification.

I called my first sequence p, pr, pr^2 and my second x, xs, xs^2 where r and s are the common ratios.

Construct expressions for the third sequence and use 't' for its common ratio.

Eliminate t and simplify. You should be able to cancel most of the algebra and end up with (r-s)^2 = 0. ..... ie. r = s

Substitute back into a 't' expression to show that t = r = s.

Question3. This has got me beaten at the moment but I'll keep at it. I've made two diagrams below that suggest you cannot even be confident about which sides include the angles that sum to 120.

What theorems / formulas have you met? eg. Do you know area of triangle = half a * b * sine C where C is the angle between a and b ?

I feel this is the way to proceed, especially as sine 120 = -sine 240 (The sum of the other two angles must be 240).

I'll post again when I've had some more inspiration.

Hope that helps a bit,

Bob

Last edited by Bob (2010-07-28 22:33:34)

Bob · 2010-07-28 22:20:10

Hi Mukesh,

Got it! The improved diagram below shows there are two cases, which lead to the same answer. With my software (Sketchpad) I've only got the accuracy set to the nearest unit so one quadrilateral gives 17 and the other 16, but with greater accuracy they would be the same and equal to your required answer.

For the left hand diagram, you can calculate angle A.

Get two cosine rule expressions for BD and equate them. [BD^2 = 3^2 + 6^2 - 2 * 3 * 6 * cos A etc]

Replace angle C with 120 - A and use compound angle formula to get an equation with cos A and sinA, find cos(A+ alpha) where alpha is known and thus find A.

Then C = 120 - A.

Then use the half side x side x sine angle formula to get the area of ABD and also CBD and add them together.

You'll then have to do it all again for the second diagram.

Bob

Bob · 2010-07-29 08:44:44

Hi Mukesh,

When you joined you asked about modulus inequalities.

Do you still want help on this?

Post an example if you do.

Bob

Math Is Fun Forum

#1 2010-07-28 20:32:51

qstion on g.p. and trigonometry

#2 2010-07-28 21:39:14

Re: qstion on g.p. and trigonometry

#3 2010-07-28 22:20:10

Re: qstion on g.p. and trigonometry

#4 2010-07-29 08:44:44

Re: qstion on g.p. and trigonometry

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