Question
In $\triangle\text{ABC},$ if a= 18, b = 24 and c = 30 and $\angle\text{C}=90^{\circ},$ find $\sin\text{A},\sin\text{B}$ and $\sin\text{C}.$
✓
Answer
a = 18, b = 24, c = 30, $\angle\text{C}=90^{\circ}$
let $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}$
$\frac{\sin\text{A}}{\text{18}}=\frac{\sin\text{B}}{\text{24}}=\frac{\sin\text{90}}{\text{30}}$
$\frac{\sin\text{A}}{18}=\frac{\sin\text{B}}{24}=\frac{1}{30}$
$\frac{\sin\text{A}}{18}=\frac{1}{30}\Rightarrow\sin\text{A}=\frac{18}{30}=\frac{3}{5}$
$\frac{\sin\text{B}}{24}=\frac{1}{30}\Rightarrow\sin\text{B}=\frac{24}{30}=\frac{4}{5}$
$\therefore\sin\text{}\text{A}=\frac{3}{5},\sin\text{B}=\frac{4}{5},\sin\text{C}=1$
let $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}$
$\frac{\sin\text{A}}{\text{18}}=\frac{\sin\text{B}}{\text{24}}=\frac{\sin\text{90}}{\text{30}}$
$\frac{\sin\text{A}}{18}=\frac{\sin\text{B}}{24}=\frac{1}{30}$
$\frac{\sin\text{A}}{18}=\frac{1}{30}\Rightarrow\sin\text{A}=\frac{18}{30}=\frac{3}{5}$
$\frac{\sin\text{B}}{24}=\frac{1}{30}\Rightarrow\sin\text{B}=\frac{24}{30}=\frac{4}{5}$
$\therefore\sin\text{}\text{A}=\frac{3}{5},\sin\text{B}=\frac{4}{5},\sin\text{C}=1$
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