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gmany

[tex] \sin\theta=\dfrac{opposite}{hypotenuse}\\\\\cos\theta=\dfrac{adjacent}{hypotenuse}\\\\\tan\theta=\dfrac{opposite}{adjacent} [/tex]

It's an isosceles triangle, therefore AB = BC.

Let AB = BC = x abd AC = y. Then:

[tex]opposite=adjacent=x\\hypotenuse=y\\\\\sin A=\dfrac{x}{y}\\\\\cos A=\dfrac{x}{y}\\\\\sin C=\dfrac{x}{y}\\\\\cos C=\dfrac{x}{y}\\\\\tan C=\dfrac{x}{x}=1[/tex]

Answer: C. tan C

Answer:

correct option is d) cos C

Step-by-step explanation:

Since,

[tex]SinФ = \frac{perpendicular}{hypotenuses}[/tex]

[tex]CosФ = \frac{base}{hypotenuses}[/tex]

[tex]TanФ = \frac{perpendicular}{base}[/tex]

In triangle ABC , sides are denoted as mention in figure-1

[tex]sin A =\frac{x}{z}[/tex]

[tex]cos A =\frac{y}{z}[/tex]

[tex]sin C =\frac{y}{x}[/tex]

[tex]tan C =\frac{y}{x}[/tex]

[tex]cos C =\frac{x}{z}[/tex]

hence, cos C have similar value as sin A

Therefore, correct option is d) cos C

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