# In a triangle of base a the ratio of the other two sides is r (2

Jitender Pal
8 years ago
Hello Student,
Given that, In ∆ABC, base = a And c/b = r To find altitude, h.
We have, in ∆ABD,
h = c sin B = c a sin B/a
= c k sin A sin B/k sin A = c k sin A sin B/sin (B + C)
= c sin A sin B sin (B – C)/sin (B + C) sin (B – C) =c sin A sin B sin (B – C)/sin2 B – sin2 C
= $\frac{c.\frac{a}{k}.\frac{b}{k}sin(B-C)}{\frac{b^2}{k^2}-\frac{c^2}{k^2}}$ = abc sin (B – C)/b2 – c2
= a(c/b) sin (B –C)/1 – (c/b)2 = ar sin (B – C)/1 – r2 ≤ ar/1 – r2
[ ∵ sin (B – C) ≤ 1] ∴ h ≤ ar/1 – r2 Hence Proved.

Thanks
Jitender Pal