(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: $

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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: $AB^{2} + BC^{2} = AC^{2}$.$AB^{2} + 3^{2} = 5^{2}$$AB^{2} + 9 = 25$$AB^{2} = 25 - 9 = 16$$AB = \sqrt{16} = 4$.Now,$\tan A = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{BC}{AB}$.$\tan A = \frac{3}{4}$.

Class 10 Mathematics Introduction to Trigonometry Mix Examples - Introduction to Trigonometry

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

A
$\frac{3}{4}$
B
$\frac{5}{3}$
C
$\frac{4}{5}$
D
$\frac{3}{2}$

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

Similar Questions

If $\tan \theta = \frac{4}{3}$,then $\frac{5 \sin \theta + 2 \cos \theta}{3 \sin \theta - \cos \theta} = \ldots \ldots$

If $\tan A = \cot B$,then $A + B = \ldots$

If $\cos 9 \alpha = \sin \alpha$ and $9 \alpha < 90^{\circ},$ then the value of $\tan 5 \alpha$ is

Prove that $\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta = 1$.

Prove that,$(\sqrt{3}+ 1) (3-\cot 30^{\circ})=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$

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Prove that,
$(\sqrt{3}+ 1) (3-\cot 30^{\circ})=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$