Question

If sintheta= 11/61, find costheta and tan theta​

Answer 1

Answer:

cos theta = 60/61

tan theta = 11/60

Step-by-step explanation:

sin² theta + cos² theta = 1

cos theta = √1² - sin² theta

cos theta = √ 1 - 121/3721

cos theta = √ 3600/3721

cos theta = 60/61

tan theta = sin theta/cos theta

tan theta = 11/61 x 61/60

tan theta = 11/60

Answer 2

Given :-

sinθ = 11/61

To find :-

cosθ , tanθ

To know :-

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

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

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

Solution :-

According to above information ,

sinθ = 11/61

That means ,

opposite side = 11

Hypotenuse = 61

In a right angle triangle ATQ pythagoras theorem We can find adjacent side of a triangle

From above triangle,

Pythagoras theorem =

(opp)² + (adj)² = (hyp)²

(11)² + (adj)² = (61)²

121 + (adj)² = 3721

(adj)² = 3721 - 121

(adj)² = 3600

(adj)² = (60)²

adj = 60 cm

So, adjacent side of a triangle is 60 cm

Now,

• adjacent side = 60cm
• opposite side = 11 cm
• Hypotenuse = 61cm

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

${cos\theta} = \dfrac{60cm}{61cm}$

${cos\theta} = \dfrac{60}{61}$

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

${tan\theta} = \dfrac{11cm}{60cm}$

${tan\theta} = \dfrac{11}{60}$

Know more :-

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