In Delta ABC,m angle B = 90^{circ},BC = 3 and | vedclass

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(A) In $\Delta ABC$,$m \angle B = 90^{\circ}$,therefore $\overline{AC}$ is the hypotenuse.Given $BC = 3$ and $AC = 5$.Using the Pythagorean theorem: $

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) In $\Delta ABC$,$m \angle B = 90^{\circ}$,therefore $\overline{AC}$ is the hypotenuse.Given $BC = 3$ and $AC = 5$.Using the Pythagorean theorem: $AB^{2} + BC^{2} = AC^{2}$.$AB^{2} + 3^{2} = 5^{2}$$AB^{2} + 9 = 25$$AB^{2} = 25 - 9 = 16$$AB = \sqrt{16} = 4$.Now,$	an A = \frac{	ext{opposite side}}{	ext{adjacent side}} = \frac{BC}{AB}$.$	an A = \frac{3}{4}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$BC = 3$ and $AC = 5$,then $\tan A = \ldots$

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