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wolverine
1 month ago
6

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

Mathematics
1 answer:
Zina [12.3K]1 month 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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