Question

If p isa point on the altitude ad of triangle abc such that angle cbp=b/3 , then prove that ab = 2 sin/3

Answer 1

Answer:

AD=csinB

In △BDP

tan 3B

= BDPD

PD=tan 3B

BD=ccosBtan 3B

AD=AP+PD

csinB=AP+ccosBtan 3B

AP=csinB−ccosBtan 3B

AP= cos 3Bccos 3B

sinB−ccosBsin 3B

AP= cos 3Bcsin 32B

=2sin 3B