If sin x=

Given $\sin x=\frac{-24}{25}$ i.e $\mathrm{x}$ lies in III or IV quadrant

$\sin x=\frac{A B}{A C}=\frac{-24}{25}$

Since $A C^{2}=A B^{2}+B C^{2}$

i. e $25^{2}=(24)^{2}+B C^{2}$

$\Rightarrow 625=576+B C^{2}$

$\Rightarrow B C^{2}=49$

$B C=\pm 7$

$\Rightarrow \tan x=\frac{24}{7}$ or $\tan x=\frac{-24}{7}$

sinx=ABAC=-2425Since AC2=AB2+BC2i.e 252=242+BC2⇒625=576+BC2⇒BC2=49BC=±7⇒tan x=247 or tan x=-247">sin