(B) In $\Delta ABC$, $AC = 5$ and hypotenuse $BC = 13$, with $m \angle A = 90^\circ$.By the Pythagorean theorem:$AB^2 = BC^2 - AC^2$$AB^2 = 13^2 - 5^2

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(B) In $\Delta ABC$, $AC = 5$ and hypotenuse $BC = 13$, with $m \angle A = 90^\circ$.By the Pythagorean theorem:$AB^2 = BC^2 - AC^2$$AB^2 = 13^2 - 5^2 = 169 - 25 = 144$$AB = \sqrt{144} = 12$Now, $\tan B = \frac{\text{side opposite to } \angle B}{\text{side adjacent to } \angle B} = \frac{AC}{AB} = \frac{5}{12}$.

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

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In $\Delta ABC$, $AC = 5$, $BC = 13$, $m \angle A = 90^\circ$, then $\tan B = \ldots$

A
$\frac{5}{13}$
B
$\frac{5}{12}$
C
$\frac{12}{13}$
D
$\frac{12}{5}$

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Mix Examples - Introduction to Trigonometry

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In $\Delta ABC$, $AC = 5$, $BC = 13$, $m \angle A = 90^\circ$, then $\tan B = \ldots$

Similar Questions

If $\sin \theta + \sin^2 \theta = 1$,then $\cos^2 \theta + \cos^4 \theta = \dots$

If $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$,then $\theta = \ldots$ (in $^{\circ}$)

If $\sin \theta + \cos \theta = p$ and $\sec \theta + \operatorname{cosec} \theta = q,$ then prove that $q(p^2 - 1) = 2p$.

$\frac{\sin 60^{\circ} + \cos 30^{\circ}}{1 + \sin 30^{\circ} + \cos 60^{\circ}} = \dots$

$\sin^{2} 60^{\circ} - \tan 45^{\circ} + \cos^{2} 30^{\circ} - \cot 90^{\circ} = \ldots$

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