Ques. -30

Geometry Level 4

If $cosA+cosB+2\cdot cosC=2$ , then the sides of $\bigtriangleup ABC$ are in :

A,B,C are the angles of $\bigtriangleup ABC$ .

None of the give sequences. Arithmetic Progression Geometric Progression Harmonic Progression

2 solutions

Prajwal Kavad
Mar 18, 2015

jee main approach:

let the triangle to be a right angle triangle and evaluate cosA, cosB, cosC as ratios

obviously one of them will be 0(due to cos 90) then u will get a pure AP

An objective approach to the problem is to put A = B= C = 60. This is clearly the case of an equilateral triangle. The sides will be thus equal. So they form an AP with zero common difference ( a valid AP according to the definition).

But then it will also form a valid GP and HP. That's what confuses me right now.

Pulkit Gupta - 5 years, 8 months ago

I did the same thing.I put $60^\circ$ each.then i thought it to be in GP. But would you give me a clear conception about AP and GP.

Tanvir Hasan - 5 years, 4 months ago

On assuming a right angled triangle, the base condition ( =2 ) remains unsatisfied ( since other two angles of a triangle can not be zero).

Pulkit Gupta - 5 years, 8 months ago

Put cosC =3/5, cosA,cosB= 4/5 or 0, with 3,4,5 as sides

Raghav Rathi - 5 years, 4 months ago

Prakhar Bindal
Feb 26, 2015

Using C and D Formula 2cos(A+B/2)cos(A-B/2) = 2(1-cosC) cos(A+B/2)cos(A-B/2) = 2 sin^2 (C/2) Replace C/2 By 90-(A+B/2) cos(A+B/2)cos(A-B/2) = 2 cos^2(A+B/2) Cancel factor of cos(A+B/2) Because it cant be zero has A,B,C Are angles of triangle cos(A-B/2) = 2cos(A+B/2) cos(A-B/2) / cos(A+B/2) = 2 Apply componendo and dividendo on both sides to get 3sinA/2sinB/2 = cosA/2cosB/2 Apply formula of sines and cosines of half angles on simplification s =3s-3c where s is the semiperimeter 2s = 3c a+b+c = 3c a+b = 2c so they are in AP

Ques. -30

2 solutions

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