Question

Given that $\sf{ Sin \; \theta = \dfrac{12}{13} }$

Than, find the Value of :-

$\sf{ \dfrac{ 2 \; Cos \; \theta + 3 \; tan \; \theta }{ Sin \; \theta + tan \; \theta \; \; Sin \; \theta }}$


Answer Only Mods , Stars and Best Users :)​

Answer 1

$sin = \frac{12}{13}$
Therefore, $o = 12, h = 13$

$a^2 + o^2 = h^2$

$a = 5$

We have:

$\frac{2cos + 3tan}{sin + tansin}$

[tex]\frac{2*\frac{5}{13}+ 3* \frac{12}{5}}{\frac{12}{13} + \frac{12}{5}*\frac{12}{13}}\\\\ \frac{\frac{10}{13}+ \frac{36}{5}}{\frac{12}{13} + \frac{144}{65}}\\\\ \frac{\frac{50}{65} + \frac{468}{65}}{\frac{60}{65} + \frac{144}{65}}\\\\ \frac{518}{204}\\\\ \frac{259}{102}[/tex]

So your answer is 259/102

Hope this helped!

Answer 2

Given :

• $\sf{ Sin \; \theta = \dfrac{12}{13} }$

To Find :

• $\sf{ \dfrac{ 2 \; Cos \; \theta + 3 \; tan \; \theta }{ Sin \; \theta + tan \; \theta \; \; Sin \; \theta }}$

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SolutioN :

❍ According to the Question :

• We know That :

$\dashrightarrow \; {\underline{\boxed{\pmb{\purple{\sf{ Sin \; \theta = \dfrac{ Perpendicular }{ Hypotenuse } }}}}}}$

• Hence :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Sin \; \theta = \dfrac{ Perpendicular }{ Hypotenuse } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\sf { Sin \; \theta = \dfrac{ 12 }{ 13 } }}}} \; {\orange{\bigstar}} \\ \\ \\ \end{gathered}$

❍ Deriving the Base :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {PR}^{2} = {PQ}^{2} + {QR}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {13}^{2} = {12}^{2} + {QR}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 169 = 144 + {QR}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sqrt{169 - 144} = QR } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sqrt{25} = QR } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\sf { QR = 5 }}}} \; {\green{\bigstar}} \\ \\ \\ \end{gathered}$

❍ Deriving the Value of Equation :

• We Know That :

$\dashrightarrow \; {\underline{\boxed{\pmb{\purple{\sf{ Cos \; \theta = \dfrac{ Base }{ Hypotenuse } }}}}}}$

$\dashrightarrow \; {\underline{\boxed{\pmb{\purple{\sf{ Tan \; \theta = \dfrac{ Perpendicular }{ Base } }}}}}}$

• Hence :

$\begin{gathered} \dashrightarrow \; \; \sf { \dfrac{ 2 \; Cos \; \theta + 3 \; tan \; \theta }{ Sin \; \theta + tan \; \theta \; \; Sin \; \theta } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup 2 \times \dfrac{Base}{Hypotenuse} \bigg\rgroup + \bigg\lgroup 3 \times \dfrac{ Perpendicular }{Base} \bigg\rgroup }{ \bigg\lgroup \dfrac{Perpendicular}{Hypotenuse} \bigg\rgroup + \bigg\lgroup \dfrac{Perpendicular}{Base} \bigg\rgroup \times \bigg\lgroup \dfrac{Perpendicular}{Hypotenuse} \bigg\rgroup } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { \dfrac{ \bigg\lgroup 2 \times \dfrac{5}{13} \bigg\rgroup + \bigg\lgroup 3 \times \dfrac{ 12 }{ 5 } \bigg\rgroup }{ \bigg\lgroup \dfrac{12}{13} \bigg\rgroup + \bigg\lgroup \dfrac{12}{5} \times \dfrac{12}{13} \bigg\rgroup }} \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup \dfrac{10}{13} + \dfrac{ 36 }{ 5 } \bigg\rgroup }{ \bigg\lgroup \dfrac{12}{13} \bigg\rgroup + \bigg\lgroup \dfrac{12}{5} \times \dfrac{12}{13} \bigg\rgroup } \\ \\ \\ \end{gathered}$

• LCM of 13 and 5 = 65

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup \dfrac{50}{65} + \dfrac{ 468 }{ 65 } \bigg\rgroup }{ \bigg\lgroup \dfrac{12}{13} \bigg\rgroup + \bigg\lgroup \dfrac{12}{5} \times \dfrac{12}{13} \bigg\rgroup } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup \dfrac{518}{65} \bigg\rgroup }{ \bigg\lgroup \dfrac{12}{13} + \dfrac{144}{65} \bigg\rgroup} \\ \\ \\ \end{gathered}$

• LCM of 65 and 13 = 65

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup \dfrac{518}{65} \bigg\rgroup }{ \bigg\lgroup \dfrac{60}{65} + \dfrac{144}{65} \bigg\rgroup } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup \dfrac{518}{65} \bigg\rgroup }{ \bigg\lgroup \dfrac{204}{65} \bigg\rgroup } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf \dfrac{ \bigg\lgroup \dfrac{518}{ \cancel{65} } \bigg\rgroup }{ \bigg\lgroup \dfrac{204}{ \cancel{65} } \bigg\rgroup } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { \dfrac{518}{204} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { \cancel\dfrac{518}{204} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\sf{ \dfrac{259}{102} }}}} \; {\red{\bigstar}} \\ \\ \\ \end{gathered}$

• Therefore :

$\qquad \; {\underline{\overline{\boxed{\pmb{\sf{ \dfrac{ 2 \; Cos \; \theta + 3 \; tan \; \theta }{ Sin \; \theta + tan \; \theta \; \; Sin \; \theta} = \dfrac{259}{102} }}}}}}$

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