(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 Trigonometric Ratios

Medium

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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Introduction to Trigonometry

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Trigonometric Ratios

Given $\tan A = \frac{4}{3},$ find the other trigonometric ratios of the $\angle A$.

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If $\sin A = \frac{3}{4},$ calculate $\cos A$ and $\tan A$.

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If $\tan \theta = \frac{5}{12}$,then $\cos \theta = \ldots$

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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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In a right triangle $ABC$,right-angled at $B$. If $\tan A = 1$,then verify that $2 \sin A \cos A = 1$.

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

Similar Questions

Given $\tan A = \frac{4}{3},$ find the other trigonometric ratios of the $\angle A$.

If $\sin A = \frac{3}{4},$ calculate $\cos A$ and $\tan A$.

If $\tan \theta = \frac{5}{12}$,then $\cos \theta = \ldots$

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

In a right triangle $ABC$,right-angled at $B$. If $\tan A = 1$,then verify that $2 \sin A \cos A = 1$.

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