# In $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}$,$B C=7 \mathrm{~cm}$. Determine :

Question.

In $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}$,$B C=7 \mathrm{~cm}$. Determine :

(i) $\sin A, \cos A$

(ii) $\sin C, \cos C$.

Solution:

By Pythagoras Theorem,

$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}=(24)^{2}+(7)^{2}=625$

$\Rightarrow \mathrm{AC}=\sqrt{625}=25 \mathrm{~cm}$

(i) $\sin \mathrm{A}=\frac{\mathbf{B C}}{\mathbf{A C}}\left\{\right.$ ie, $\left.\frac{\text { sile opposite to ang } \mathbf{A}}{\text { Hyp. }}\right\}$

$=\frac{\mathbf{7}}{\mathbf{2 5}}(\because \mathrm{BC}=7 \mathrm{~cm}$ and $\mathrm{AC}=25 \mathrm{~cm})$

$\cos \mathrm{A}=\frac{\mathbf{A B}}{\mathbf{A C}}\left\{\right.$ ie, $\left.\frac{\text { sile are acent to angle } \mathbf{A}}{\text { Hyp. }}\right\}$

$=\frac{24}{25}(\because \mathrm{AB}=24 \mathrm{~cm}$ and $\mathrm{AC}=25 \mathrm{~cm})$

(ii) $\sin \mathrm{C}=\frac{\mathbf{A B}}{\mathbf{A C}}\left\{\right.$ ie, $\left.\frac{\text { side opposte to angle } \mathbf{C}}{\text { Hyp. }}\right\}$

$=\frac{24}{25}$

$\cos \mathrm{C}=\frac{\mathbf{B C}}{\mathbf{A C}}\left\{\right.$ ie, $\left.\frac{\text { sile argicent to angle } \mathbf{C}}{\text { Hyp. }}\right\}$

$=\frac{7}{25}$