Question

given cotA =root 1/3 find all other trigonometric ratios​


Don't answer if you can't...​

Answer 1

Given:-

• CotA = $\sf{\dfrac{1}{\sqrt{3}}}$

To find:-

All other trigonometric ratios.

Solution:-

We know,

CotA = $\sf{\dfrac{Base}{Perpendicular}}$

Therefore,

$\sf{\dfrac{1}{\sqrt{3}} = \dfrac{Base}{Perpendicular}}$

Now,

We have,

Base = 1 units

Perpendicular = √3 units.

Now using pythagoras theorem we can find out hypotenuse.

According to Pythagoras Theorem

$\sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}$

= $\sf{Hypotenuse = \sqrt{(1)^2 + (\sqrt{3})^2}}$

= $\sf{Hypotenuse = \sqrt{1+3}}$

= $\sf{Hypotenuse = \sqrt{4}}$

= $\sf{Hypotenuse = 2\:units}$

Now,

Let us find all the other trigonometric ratios.

For SinA

$\sf{SinA = \dfrac{Perpendicular}{Hypotenuse}}$

= $\sf{SinA = \dfrac{\sqrt{3}}{2}}$

For CosA

$\sf{CosA = \dfrac{Base}{Hypotenuse}}$

= $\sf{CosA = \dfrac{1}{2}}$

For TanA

$\sf{TanA = \dfrac{Perpendicular}{Base}}$

= $\sf{TanA = \dfrac{\sqrt{3}}{1} = \sqrt{3}}$

For CosecA

$\sf{CosecA = \dfrac{Hypotenuse}{Perpendicular}}$

= $\sf{CosecA = \dfrac{2}{\sqrt{3}}}$

For SecA

$\sf{SecA = \dfrac{Hypotenuse}{Base}}$

= $\sf{SecA = \dfrac{2}{1} = 2}$

For CotA

$\sf{CotA = \dfrac{Base}{Perpendicular}}$

= $\sf{CotA = \dfrac{1}{\sqrt{3}}}$

Therefore all the trigonometric ratios are as follows:-

$\sf{SinA = \dfrac{\sqrt{3}}{2}}$

$\sf{CosA = \dfrac{1}{2}}$

$\sf{TanA = \sqrt{3}}$

$\sf{CosecA = \dfrac{2}{\sqrt{3}}}$

$\sf{SecA = 2}$

$\sf{CotA = \dfrac{1}{\sqrt{3}}}$

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Points To Remember!!!

• $\sf{Sin\theta = \dfrac{Perpendicular}{Hypotenuse}}$

• $\sf{Cos\theta = \dfrac{Base}{Perpendicular}}$

• $\sf{Tan\theta = \dfrac{Perpendicular}{Base}}$

• $\sf{Cosec\theta = \dfrac{Hypotenuse}{Perpendicular}}$

• $\sf{Sec\theta = \dfrac{Hypotenuse}{Base}}$

• $\sf{Cot\theta = \dfrac{Base}{Perpendicular}}$

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Answer 2

Given :

$\bf cotA = \sqrt{ \dfrac{1}{3}} </p><p></p><p> [tex]\bf \implies cotA = \dfrac{1}{\sqrt{3}}$

To Find :

All other trigonometric ratios.

sinA

cosA

tanA

cosecA

secA

Solution :

To do this sum more easily let's make a diagram of right angled triangle ABC.

Now, From point A,

AC = Hypotenuse

BC = Perpendicular

AB = Base

$\bf We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}$

$\bf We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}$

$\bf \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}$

$\bf \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}$

$\bf \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)$

$Now, by Pythagoras' theorem, we have</p><p></p><p>AC² = AB² + BC²$

$\bf \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}$

$\bf \implies AC^{2} = 1x^{2} + 3x^{2}$

$\bf \implies AC^{2} = 4x^{2}$

$\bf \implies AC = \sqrt{4x^{2}}$

$\bf \implies AC = 2x$

Now,

$\bf sin \theta = \dfrac{perpendicular}{hypotenuse}$

$\bf\implies sinA = \dfrac{BC}{AC}$

$\bf \implies sinA = \dfrac{\sqrt{3}x}{2x}$

$\bf \implies sinA = \dfrac{\sqrt{3}}{2}$

$\Large \boxed{\bf sinA = \dfrac{\sqrt{3}}{2}}$

$\bf cos \theta = \dfrac{base}{hypotenuse}$

$\bf \implies cosA = \dfrac{AB}{AC}$

$\bf \implies cosA = \dfrac{1x}{2x}$

$\bf \implies cosA = \dfrac{1}{2}$

$\Large \boxed{\bf cosA = \dfrac{1}{2}}$

$\bf tan \theta = \dfrac{perpendicular}{base}$

$\bf \implies tanA = \dfrac{BC}{AB}$

$\bf \implies tanA = \dfrac{\sqrt{3}x}{1x}$

$\bf \implies tanA = \sqrt{3}$

$\Large \boxed{\bf tanA = \sqrt{3}}$

$\bf cosec \theta = \dfrac{hypotenuse}{perpendicular}$

$\bf \implies cosecA = \dfrac{AC}{BC}$

$\bf \implies cosecA = \dfrac{2x}{\sqrt{3}x}$

$\bf \implies cosecA = \dfrac{2}{\sqrt{3}}$

$\Large \boxed{\bf cosecA = \dfrac{2}{\sqrt{3}}}$

$\bf sec \theta = \dfrac{hypotenuse}{base}$

$\bf \implies secA = \dfrac{AC}{AB}$

$\bf \implies secA = \dfrac{2x}{1x}$

$\bf \implies secA = 2$

$\Large \boxed{\bf secA = 2}$