Question

If cos theta =12÷13 find tan theta

Answer 1

Given :-

$\sf{cos\theta} = \sf\dfrac{12}{13}$

To find :-

$\sf{tan\theta}$

SoLuTiOn :-

As we know that ,

$\sf{cos\theta} = \dfrac{adjacent}{hypotenuse}$

Since ,

• Adjacent side = 12
• Hypotenuse = 13

From Pythagoras theorem we can find opposite side

$\sf{(Adj)^2 +(opp)^2 = (hyp)^2}$

$\sf{ (12)^2 + (opp)^2 = (13)^2}$

$\sf{ 144 + (opp)^2 = 169}$

$\sf{(opp)^2 = 169- 144}$

$\sf{(opp)^2 = 25}$

$\sf{(opp)^2 = 5^2}$

$\sf{opp = 5}$

Since opposite side is 5

• opposite side- 5
• hypotenuse - 13
• adjacent- 12

Now ,

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

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

Required answer:-

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

Know more :-

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

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