Question

anawer the following ​

Answer 1

Solution:

Step-by-step explanation:

Given :

$\large sin \ A=\dfrac{1}{3}$

We know that

$\large sin \ A=\dfrac{Perpendicular}{hypotenuse}$

We have pythagoras theorem

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

Putting values here we get

$\large 3^2=1^2+B^2\\\\\\\large 9=1+B^2\\\\\\\large B=\sqrt{9-1}\\\\\\\large B=\sqrt{8}\\\\\\\large B=2\sqrt{2}$

Now for other ratio equal to

$\large cos \ A=\dfrac{base}{hypotenuse}=\dfrac{2\sqrt2}{3}\\\\\\\large tan \ A=\dfrac{perpendicular}{base}=\dfrac{1}{2\sqrt2}=\dfrac{2\sqrt2}{8} \\\\\\\large cot \ A=\dfrac{base}{perpendicular}=\dfrac{2\sqrt2}{1}=2\sqrt2\\\\\\\large sec \ A=\dfrac{hypotenuse}{base}=\dfrac{3}{2\sqrt2}=\dfrac{6\sqrt2}{8} \\\\\\\large cosec \ A=\dfrac{hypotenuse}{perpendicular}=\dfrac{3}{1}=3$

Thus we get answer.