Question

PLEASE ANSWER WILL GIVE POINTS!!!!!!!!!!!

Answer 1

Given:

The figure of a right triangle ABC.

To find:

The trigonometric ratios $\sin B, \tan A, \cos B$.

Solution:

Using Pythagoras theorem,

$Hypotenuse^2=Perpendicular^2+Base^2$

$AB^2=AC^2+BC^2$

$(10)^2=(6)^2+BC^2$

$100-36=BC^2$

$64=BC^2$

Taking square root on both sides.

$\sqrt{64}=BC$

$8=BC$

So, measure of BC is 8 units.

Now,

$\sin \theta=\dfrac{Opposite}{Hypotenuse}$

$\sin B=\dfrac{AC}{AB}$

$\sin B=\dfrac{6}{10}$

$\sin B=\dfrac{3}{5}$

Similarly,

$\tan \theta=\dfrac{Opposite}{Adjacent}$

$\tan A=\dfrac{AC}{BC}$

$\tan A=\dfrac{6}{8}$

$\tan A=\dfrac{3}{4}$

And,

$\cos \theta=\dfrac{Adjacent}{Hypotenuse}$

$\cos B=\dfrac{BC}{AB}$

$\cos B=\dfrac{8}{10}$

$\cos B=\dfrac{4}{5}$

Therefore, the required trigonometric ration are $\sin B=\dfrac{3}{5}, \tan A=\dfrac{3}{4}, \cos B=\dfrac{4}{5}$.