If in $△ A B C$ , $∠ B$ is right angle and AB=12cm and BC=5cm, then find the value of $sin A, tan A, sin C$ and $cot C$

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Solution

$\begin{matrix} A C & = \sqrt{12^{2} + 5^{2}} = 13 thus & sin A = \frac{B C}{A C} = \frac{5}{13} \end{matrix}$
$\begin{matrix} tan A = \frac{B C}{A B} = \frac{5}{12} sin C = \frac{A B}{A C} = \frac{12}{13} cot C = \frac{B C}{A B} = \frac{5}{12} \end{matrix}$

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If in △ABC,∠B is right angle and AB=12cm and BC=5cm, then find the value of sinA,tanA,sinC and cotC

AC=√122+52=13 thus sinA=BCAC=513tanA=BCAB=512sinC=ABAC=1213cotC=BCAB=512

If in $△ A B C$ , $∠ B$ is right angle and AB=12cm and BC=5cm, then find the value of $sin A, tan A, sin C$ and $cot C$

$\begin{matrix} A C & = \sqrt{12^{2} + 5^{2}} = 13 thus & sin A = \frac{B C}{A C} = \frac{5}{13} \end{matrix}$
$\begin{matrix} tan A = \frac{B C}{A B} = \frac{5}{12} sin C = \frac{A B}{A C} = \frac{12}{13} cot C = \frac{B C}{A B} = \frac{5}{12} \end{matrix}$