Question

If tan A  =  2/3, then find all the other trigonometric ratios. ​

Answer 1

Solution :

tan A = opposite side / adjacent side = 2 / 3

By Pythagorean Theorem,

AC2 = AB2 + BC2

AC2 = 32 + 22

AC2 = 9 + 4

AC2 = 13

AC = √13

Then,

sin A = opposite side / hypotenuse = BC/ AC = 2/√13

cos A = adjacent side / hypotenuse = AB/AC = 3/√13

csc A = 1 / sin A = √13/2

sec A = 1 / cos A = √13/3

cot A = 1 / tan A = 3/2

Answer 2

Answer:

All the other trigonometric ratios are:

$sinA=\frac{2}{\sqrt{13} }$, $cosA=\frac{3}{\sqrt{13} }$, $secA=\frac{\sqrt{13} }{3}$, $cosecA=\frac{\sqrt{13} }{2}$ and $cotA=\frac{3 }{2}$.

Step-by-step explanation:

Step 1 of 6

Given: $tanA=\frac{2}{3}$

We know that,

$tanA=\frac{perpendicular\ side\ to\ angle\ A}{base \ side\ of\ angle\ A}$

$tanA=\frac{BC}{AB}$

So, in the figure,

Δ$ABC$ is a right angled triangle in which $AB=3$ units and $BC=2$ units.

Then by Pythagoras theorem,

$AC=\sqrt{AB^2+BC^2}$

      $=\sqrt{3^2+2^2}$

      $=\sqrt{9+4}$

$AC=\sqrt{13}$ units

Step 2 of 6

Computing the trigonometric ratio of $sinA$.

$sin A=\frac{BC}{AC}$

$sinA=\frac{2}{\sqrt{13} }$

Step 3 of 6

Computing the trigonometric ratio of $cosA$.

$cos A=\frac{AB}{AC}$

$cosA=\frac{3}{\sqrt{13} }$

Step 4 of 6

Computing the trigonometric ratio of $secA$.

$sec A=\frac{1}{cosA}$

        $=\frac{1}{3/\sqrt{13} }$

$secA=\frac{\sqrt{13} }{3}$

Step 5 of 6

Computing the trigonometric ratio of $cosecA$.

$cosec A=\frac{1}{sinA}$

        $=\frac{1}{2/\sqrt{13} }$

$cosecA=\frac{\sqrt{13} }{2}$

Step 6 of 6

Computing the trigonometric ratio of $cotA$.

$cot A=\frac{1}{tanA}$

        $=\frac{1}{2/3 }$

$cotA=\frac{3 }{2}$

Final answer: all the other trigonometric ratios are:

$sinA=\frac{2}{\sqrt{13} }$, $cosA=\frac{3}{\sqrt{13} }$, $secA=\frac{\sqrt{13} }{3}$, $cosecA=\frac{\sqrt{13} }{2}$ and $cotA=\frac{3 }{2}$.

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