2010/014) Let a,b and c be the sides of a right angled triangle. Let theta be the smallest angle of this triangle.?

If 1/a, 1/b, 1/c are also the sides of a right angled triangle then show that Sin(theta) = (sqrt(5) - 1)/2;

proof
Let a < b < c


And theta (say t opposite to smaller side)

Sin t = a/c and cos t = b/c

Now 1/a > 1/b > 1/c and as it is right angled triangle we have

(1/a)^2 = (1/b)^2 + (1/c)^2

Or

(c/a)^2 = (c/b)^2 +1

(1/sin ^2 t) = (1/ cos^2 t ) +1

Or cos^2 t = sin ^2 t + sin ^2 t cos^2 t

Or 1- sin ^2t = sin ^2 t + sin ^2 t (1- sin ^2 t)
= 2 sin ^2 t – sin ^4 t
Or sin ^4 t – 3 sin ^2 t + 1 = 0

Sin^2 t = (3 +/- sqrt(9-4))/2 = (3 – sqrt(5))/2 , + cannot be taken as it shall be >1 not possible


So sin t = sqrt((3 – sqrt(5))/2) say sqrt(x) – sqrt(y) as there is sqrt(5) in square

So squaring we get

x+y – 2sqrt(ab) = (3 –sqrt(5))/2

so x+y = 3/2 and sqrt(xy) = sqrt(5)/4) or xy = 5/16

we can solve it as x = 5/4 and y= 1/4 or y = 1/4 and x = 5/4

so sin t = +/-(sqrt(5/4) - sqrt(1/4))

positive value to be taken as sin t > 0 so sin t = sqrt(5/4) – sqrt(1/4) = (sqrt(5)-1)/2