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10 days ago

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

Mathematics

1 answer:

Zina [12.3K]10 days ago

8 0

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:

Provided that

In a right angle triangle ΔABC, ∠C =90°.

The scenario is illustrated below,

In triangle ΔABC:-

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

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

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

Now, in the case of ∠A, the trigonometric ratio dimensions will vary.

Here, the base for ∠A is AC, the perpendicular side is CB, while the hypotenuse remains consistent across ratios

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

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

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

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

Therefore,

All trigonometric ratios equal $SinB = \frac{AC}{AB}$ , $SinA= \frac{CB}{AB}$ , $CosA= \frac{AC}{AB}$

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

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