Question

IMPORTANT PLEASE HELP
Explain the difference between using the cosine ratio to solve for a missing angle in a right triangle versus using the secant ratio. You must use complete sentences and any evidence needed (such as an example) to prove your point of view

Answer 1

Answer:

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Answer 2

Answer with Step-by-step explanation:

Suppose a right triangle ABC in which

BC=3 units and AC=5 unit and angle B=90 degree

Let x be the missing angle of right triangle

We know that

$cos x=\frac{base}{hypotenuse}$

$sec x=\frac{hypotenuse}{base}$

$cosx=\frac{3}{5}=0.6$

$x=cos^{-1}(0.6)=53.1^{\circ}$

$sec x=\frac{1}{cos x}=\frac{1}{0.6}$

$cos x=0.6$

$x=cos^{-1}(0.6)=53.1$

Hence, there is no difference between using cosine ratio to solve for a missing angle in a right triangle versus using the secant ratio.

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