In Delta ABC,mangle A = 90^circ,AB = 5,AC = 1 | vedclass

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Question

(D) In a right-angled triangle $\Delta ABC$ where $\angle A = 90^\circ$:For angle $C$,the opposite side is $AB = 5$ and the adjacent side is $AC = 12$

Vedclass · Accepted Answer

$\frac{17}{13}$

Vedclass · Answer

(D) In a right-angled triangle $\Delta ABC$ where $\angle A = 90^\circ$:For angle $C$,the opposite side is $AB = 5$ and the adjacent side is $AC = 12$. The hypotenuse is $BC = 13$.Using the trigonometric ratios:$\sin C = \frac{	ext{opposite side}}{	ext{hypotenuse}} = \frac{AB}{BC} = \frac{5}{13}$$\cos C = \frac{	ext{adjacent side}}{	ext{hypotenuse}} = \frac{AC}{BC} = \frac{12}{13}$Therefore,$\sin C + \cos C = \frac{5}{13} + \frac{12}{13} = \frac{5 + 12}{13} = \frac{17}{13}$.

In $\Delta ABC$,$m\angle A = 90^\circ$,$AB = 5$,$AC = 12$ and $BC = 13$. Therefore,$\sin C + \cos C = \ldots$

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