Question

find tan theta if cos theta is πr²= -12/13​

Answer 1

Correct question :–

• Find tan(θ) , If cos(θ) = -12/13.

ANSWER :–

GIVEN :–

$\\ \bf \implies \cos( \theta) = - \dfrac{12}{13}$

TO FIND :–

$\\\implies \bf \tan( \theta) = ?$

SOLUTION :–

$\\ \bf \implies \cos( \theta) = - \dfrac{12}{13}$

• We know that –

$\\ \bf \implies \sin^{2} ( \theta) + \cos^{2} ( \theta) = 1$

$\\ \bf \implies \sin^{2} ( \theta) = 1 - \cos^{2} ( \theta)$

$\\ \bf \implies \sin( \theta) = \sqrt{1 - \cos^{2} ( \theta)}$

• Put the value of cos(θ) –

$\\ \bf \implies \sin( \theta) = \sqrt{1 - \bigg( - \dfrac{12}{13} \bigg)^{2}}$

$\\ \bf \implies \sin( \theta) = \sqrt{1 - \dfrac{144}{169}}$

$\\ \bf \implies \sin( \theta) = \sqrt{\dfrac{169 - 144}{169}}$

$\\ \bf \implies \sin( \theta) = \sqrt{\dfrac{25}{169}}$

$\\ \bf \implies \sin( \theta) = \dfrac{5}{13}$

• We also know that –

$\\\implies \bf \tan( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) }$

• So that –

$\\\implies \bf \tan( \theta) = \dfrac{ \dfrac{5}{13} }{ - \dfrac{12}{13}}$

$\\\implies \bf \tan( \theta) = \dfrac{ 5 }{ - 12}$

$\\ \large\implies{ \boxed{ \bf \tan( \theta) = - \dfrac{5}{12}}}$

Answer 2

Answer:

Given :

• Cos = -12 / 13

To Find :

• what is the value of tan

Solution :

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf -12 }\put(2.8,.3){\large\bf 5 cm}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\put(4,2.7){\large\bf 13 cm}\end{picture}$

Pythagoras theory's :

$: \implies \sf \: \: \: \: \: \: \: \: \: \: {AB}^{2} = {BC}^{2} + {AC}^{2}$

Substitute values :

$: \implies \sf \: \: \: \: \: \: \: \: \: \: {(- 12)}^{2} = {BC}^{2} + {(13)}^{2} \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \: \: \:144 = {BC}^{2} + 169 \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \: \: \: {BC}^{2} = 169 - 144 \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \: \: \: {BC}^{2} = \: 25 \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \: \: \: BC = \: 5$

$\\ \\: \implies \sf \: \: \: \: \: \: \: \: \: \: \: Tan \: = \frac{prependicular}{ base} \:$

Substitute values :

$\\ \\: \implies \sf \: \: \: \: \: \: \: \: \: \: \: Tan \: = \frac{5}{ - 12} \:$

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