Question

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

Answer 1

Answer+Step-by-step explanation:

cotA =root 1/3 = 1/root3

so , A = cot^-1( 1/root3) = cot^-1(cot60) = 60

so, A = 60 degree

now  

SinA = sin60 = root3 /2 = 0.866

CosA=cos60 = 1/2 = 0.5

TanA= tan60 = root3 = 1.732

SecA = sec60 = 2

CosecA = cosec60 = 2/root3 = 2/1.732 = 1.154

CotA = cot60 = 1/root3 = 0.577

Answer 2

Given :

$\bf cotA = \sqrt{ \dfrac{1}{3}}$

$\bf \implies cotA = \dfrac{1}{\sqrt{3}}$

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

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,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf \implies secA = 2$

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