Question

If 5 sin A = 8cos A then the value of cot A is?

Answer 1

Given :

• 5 sin A = 8cos A

To find :

• the value of CotA

Concept used :

$\dfrac{ \sin(x) }{ \cos(x) } = \tan(x)$

$\tan(x) = \dfrac{1}{ \cot(x) }$

Solution :

$\implies 5 \sin(A) = 8 \cos(A)$

$\implies 5 \times \dfrac{ \sin(A) }{ \cos(A) } = 8$

$\implies \dfrac{ \sin(A) }{ \cos(A) } = \dfrac{8}{5}$

We know that :-

$\dfrac{ \sin(x) }{ \cos(x) } = \tan(x)$

Therefore :-

$\implies { \tan(A) } = \dfrac{8}{5}$

we know that :-

$\tan(x) = \dfrac{1}{ \cot(x) }$

$\implies { \cot(A) } = \dfrac{5}{8}$

Answer :

${ \cot(A) } = \dfrac{5}{8}$

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$\large\bf\blue{Additional-Information}$

1. Cosθ = base / hypotenuse

2. cossecθ = 1/ sinθ

3. sec θ = 1/cosθ

4. Cotθ = 1/ tanθ

5. Sin²θ+ Cos²θ= 1

6. Sec²θ - tan²θ = 1

7. cosec ²θ - cot²θ = 1

8. sin(90°−θ) = cos θ

9. cos(90°−θ) = sin θ

10. tan(90°−θ) = cot θ

11. cot(90°−θ) = tan θ

12. sec(90°−θ) = cosec θ

13. cosec(90°−θ) = sec θ

14. Sin2θ = 2 sinθ cosθ

15. cos2θ = Cos²θ- Sin²θ

$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$