Question

if 5tan teta =12find the values of cos teta and sin teta​

Answer 1

Given :-

${5tan\theta} = 12$

To find :-

${cos\theta, sin\theta}$

SOLUTION:-

As they given ,

${5tan\theta} = 12$ then ,

${tan\theta}= \dfrac{12}{5}$

As we know that ,

${tan\theta} = \dfrac{opposite}{adjacent}$

So, ATQ

${opposite\: side = 12}$

${adjacent \: side= 5}$

From Pythagoras theorem,

We can find hypotenuse

Pythagoras theorem

$(opposite \: side) {}^{2} + (adjacent \: \: side) {}^{2} = (hypotenuse){}^2$

$(12) {}^{2} + (5) {}^{2} = hyp {}^{2}$

$144 + 25 = hyp {}^{2}$

$169 = hyp {}^{2}$

$13 {}^{2} = hyp {}^{2}$

$hyp = 13$

So, hypotenuse = 13

As we know that ,

${cos\theta}= \dfrac{Adjacent}{Hypotenuse}$

$☆{cos\theta} = \dfrac{5}{13}$

${sin\theta} = \dfrac{Opposite}{Hypotenuse}$

$☆{sin\theta} = \dfrac{12}{13}$

Know more :-

Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$