Question

Given : $\cos A=0.6$; trigonometrical ratio sec$A$ for angle $A$ is $\sec A=\frac{m}{3}$, m is

Answer 1

$\cos A=0.6$

$\cos A=\frac{6}{10}$

$\cos A=\frac{B}{H}=\frac{3}{5}$

Now, $B=3,H=5$

$H^2=P^2+B^2$

$5^2=3^2+P^2$

$25=9+p^2$

$P=4$

Now, $\sin A=\frac{P}{H}=\frac{4}{5}=0.8$

$\tan A=\frac{P}{B}=\frac{4}{3}=1\frac{1}{3}$

$\csc A=\frac{H}{P}=\frac{5}{4}=1\frac{1}{4}$

$\sec A=\frac{H}{B}=\frac{5}{3}=1\frac{2}{3}$

$\cot A=\frac{B}{P}=\frac{3}{4}=0.75$