Question

In right triangle ABC, mC - 90° and AC BC. Which trigonometric ratio is cquivalent to sin b?

Answer 1

Answer:

All trigonometric Ratios are  $SinB = \frac{AC}{AB}$ ,  $SinA= \frac{CB}{AB}$ ,  $CosA= \frac{AC}{AB}$

And  $Cos B = \frac{CB}{AB}$.

Step-by-step explanation:

Given that,

A right angle triangle ΔABC, ∠C =90°.

Diagram of the given scenario shown below,

In triangle ΔABC :-

                               $Hypotenuse = AB\\Base = CB\\Perpendicular = AC$

So,                            $Sin\theta = \frac{perpendicular}{hypotenuse}$

                                $SinB = \frac{AC}{AB}$

Now, for ∠A the dimensions of trigonometric ratios will be changed.

Here the base for ∠A is AC , perpendicular side is CB and hypotenuse will be same for all ratios.

                               $SinA= \frac{CB}{AB}$

Again,                    $Cos\theta= \frac{base}{hypotenuse}$

Then,                    $CosA= \frac{AC}{AB}$

And                      $Cos B = \frac{CB}{AB}$.

Hence,

All trigonometric Ratios are  $SinB = \frac{AC}{AB}$ ,  $SinA= \frac{CB}{AB}$ ,  $CosA= \frac{AC}{AB}$

And                      $Cos B = \frac{CB}{AB}$.