Question

SinA=1/root3 Find cotA

Answer 1

★ Given:

• sinA = $\sf\dfrac{1}{\sqrt{3}}$

★ To Find:

• cotA = ?

★ Diagram

$\setlength{\unitlength} {1mm}\begin{picture} (5,5)\put(0,0){\line(1,0) {30}}\put(30,0){\line(0,1) {40}}\put(0,0){\line(3,4) {30}}\put(13,-4){1.41\ cm} \put(32,17){1\ cm}\put(8,23) {1.78cm}\put(3,1){ \theta } \end{picture}$

★ Solution:

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

That means

Opposite side of $\theta = 1$

Hypotenuse of ∆ = $\sqrt{3}$

Finding adjacent side of $\theta$ by Pythagoras theorem

♦ (hypotenuse)² = (base)² + (Height)²

→ (√3)² = 1² + H²

→ H² = 3 - 1

→ H = √2

Now, cotA

cotA = $\sf \dfrac{adjacent}{opposite} = \dfrac{\sqrt{2}}{1}$

cotA = √2. [answer]

Answer 2

AnsWer :

√2.

SolutioN :

Diagram.

$\setlength{\unitlength}{1mm}}\begin{picture}(5,5)\put(0,0){\line(1,0){30}}\put(30,0){\line(0,1){40}}\put(0,0){\line(3,4){30}} \put(-3,0){ \tt{A} }\put(31,0){ \tt{B} }\put(30,40.5){ \tt{C} } \put(33,15){ \tt{1} } \put(2,15){ \tt{\sqrt{3}} }\put(12,-5){ \tt{\sqrt{2}} }\put(23,5.2){ \tt{90} }\qbezier(23,0)(26,5.5)(29.5,4.3)\end{picture}$

We have,

$\tt \dagger \: \: \: \: \: Sin\, \theta = \dfrac{1}{ \sqrt{3} }$

We know,

• Sin²A + Cos²A = 1.

$\tt \leadsto { \Bigg(\dfrac{1}{ \sqrt{3} }\Bigg)}^{2} + {cos\, }^{2} \theta =1 .$

$\tt \leadsto { \Bigg(\dfrac{1}{ 3 }\Bigg)} + {cos\, }^{2} \theta =1 .$

$\tt \leadsto {cos\, }^{2} \theta =1 - \dfrac{1}{3}$

$\tt \leadsto {cos\, }^{2} \theta = \dfrac{3 - 1}{3}$

$\tt \leadsto {cos\, }^{2} \theta = \dfrac{2}{3}$

$\tt \leadsto cos\, \theta = \pm \sqrt{ \dfrac{2}{3} }$

Now,

• P = 1.
• B = √2.
• H = √3.

$\tt \dagger \: \: \: \: \: Sin \,\theta = \dfrac{Perpendicular }{Hypotenuse }$

$\tt \dagger \: \: \: \: \: Sin \,\theta = \dfrac{1 }{ \sqrt{3} }$

$\rule{90}1$

$\tt \dagger \: \: \: \: \: Cos \,\theta = \dfrac{ Base }{Hypotenuse }$

$\tt \dagger \: \: \: \: \: Cos \,\theta = \dfrac{ \sqrt{2} }{ \sqrt{3} } = \sqrt{ \dfrac{2}{3} }$

$\rule{90}1$

$\tt \dagger \: \: \: \: \: Tan \,\theta = \dfrac{ Perpendicular }{Base }$

$\tt \dagger \: \: \: \: \: Tan \,\theta = \dfrac{ 1 }{ \sqrt{2} }$

$\rule{90}1$

$\tt \dagger \: \: \: \: \: Cot \,\theta = \dfrac{ Base }{ Perpendicular}$

$\tt \dagger \: \: \: \: \: Cot \,\theta = \dfrac{ \sqrt{2} }{ 1}$