Question

In any triangle ABC sinA-cosB=cosC then angle B =pi/2​

Answer 1

in any triangle ABC,

sinA - cosB = cosC

or, sinA = cosC + cosB

or, sinA = 2cos(B + C)/2. cos(B - C)/2

we know, A + B + C = π......(1)

so, (B + C)/2 = (π/2 - A/2)

so, cos(B + C)/2 = cos(π/2 - A/2) = sinA/2

now, sinA = 2sinA/2 .cos(B - C)/2

or, 2sinA.cosA/2 = 2sinA/2 .cos(B - C)/2

[ from sin2x = 2sinx.cosx ]

or, cosA/2 = cos(B - C)/2

or, A/2 = (B - C)/2

or, A + C = B ........(2)

from equations (1) and (2),

2B = π/ ⇒B = π/2

hence proved