A little trig

mercury

this is a question that was part of an mcq test.
and i did'nt have any clue as to how to begin!
finally i just tried plugging in values for A,B,C to see if they worked.
for ex. i put A=B=C=60 (degrees)- equilateral and so on..but i'd like to know the logical way to do it...

if A,B,C are the vertices of a triangle,
and if cosBcosC + sinAsinBsinC = 1

is the triangle
a) equilateral
b) isosceles
c) right angled isosceles
d) right angled but not isosceles
e) none of the above.

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Mulder

Right, I'm assuming you know sin^2 +cos^2 = 1.

Well you can see this is nearly similar, and almost exactly the same if B = C. So there's two angles the same.

So now cos^2 + sinA sin^2 = 1

Which means sinA must be 1.

Therefore A can only take value of 90 degrees - right angle (450...etc useless here)

180 - 90..other two angles combine as total 90 degrees, so they must be 45 each since you've already found they're the same. Which leaves you a right angled isosceles.

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