Can I use proportion to my advantage?

By ratio and proportion we know that

$\frac { a }{ b } =\frac { c }{ d } =\frac { a+c }{ b+d }$

Therefore,

$\frac { \sin ^{ 2 }{ \theta } }{ 5 } =\frac { \cos ^{ 2 }{ \theta } }{ 6 } =\frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ 11 }$

$\Rightarrow \frac { \sin ^{ 2 }{ \theta } }{ 5 } =\frac { 1 }{ 11 }$

$\Rightarrow \sin { \theta } =\pm \sqrt { \frac { 5 }{ 11 } }$

As $\theta$ is in first quadrant we have $\sin { \theta } =\sqrt { \frac { 5 }{ 11 } } =0.674$

$\large \because \ \ \ \frac{ \sin^{2}{\theta}}{5} \ = \ \frac{ \cos^{2}{\theta}}{6}$ $\large \therefore \ \ \ \frac{6}{5} \ = \ \frac{\cos^{2}{\theta}}{\sin^{2}{\theta}}$ And by adding $1$ (which is equal to $\large \frac{5}{5}$ and also $\large \frac{ \sin^{2}{\theta}}{ \sin^{2}{\theta}}$ ) to both sides , we get : $\large \frac{5+6}{5} \ = \ \frac{\cos^{2}{\theta} + \sin^{2}{\theta}}{\sin^{2}{\theta}}$ $\large \therefore \ \ \ \frac{11}{5} \ = \ \frac{1}{\sin^{2}{\theta}}$ $\large \therefore \ \ \ \sin^{2}{\theta} \ = \ \frac{5}{11}$ $\large \therefore \ \ \ \sin{\theta} \ = \ \pm \sqrt{\frac{5}{11}} \ = \ \pm \frac{\sqrt{5}}{\sqrt{11}}$ But as $\large \theta$ is a positive acute angle which lies in the first quadrant , therefore $\large \sin{\theta}$ must be positive too . $\large \therefore \ \ \ \sin{\theta} \ = \ \frac{\sqrt{5}}{\sqrt{11}} \ \approx \ \boxed{0.674}$

$\frac {(\sin\theta)^{2}}{5} = \frac {(\cos\theta)^{2}}{6} 》》》6 (\sin\theta)^{2}=5 (\cos\theta)^{2}$ Add $(5\sin\theta)^{2}$ on both sides so $11 (\sin\theta)^{2} = 5 ((\cos\theta)^{2} + (\sin\theta)^{2})》》》 11 (\sin\theta)^{2}= 5 》》》 (\sin\theta)^{2} = \frac{5}{11}》》》 \sin\theta= \sqrt {\frac {5}{11}}$

I thought of this differently to others apparently. Here's my working: $\frac { { \sin ^{ 2 }{ \theta } } }{ 5 } =\frac { { \cos ^{ 2 }{ \theta } } }{ 6 } \\ \frac { { \sin ^{ 2 }{ \theta } } }{ \cos ^{ 2 }{ \theta } } =\frac { { 5 } }{ 6 } \\ \tan ^{ 2 }{ \theta } =\frac { { 5 } }{ 6 } \\ \tan \theta =\sqrt { \frac { { 5 } }{ 6 } } \\ \theta \approx 0.7399...\\ \sin \theta =0.6742...\\ Answer=\boxed { 0.674 } \quad (to\quad 3dp)$ So, using just the identity that sine divided by cosine = tangent.

cos^2(x)+sin^2(x)=1, cos^2(x)=1-sin^2(x) , and the rest is simple :)

Suppose, Sin^2@/5=Cos^2@/6=k Then,we got sin^2@=5k and cos^2@=6K Then ,we got 2 equations 1. Sin^2@-5k=0 2.Cos^2@-6k=0 Then we replace the value of sin^2@ to equation no. 2 Then, 1-sin^2@=6K [cos^2@ = 1-sin^2@] 》1-5k=6k 》11k=1 》k=1/11 We replace the value to equation no .1 Sin^2@= 5/11 》Sin@=(5/11)^(1/2) 》Sin@=0.674

Cos^2(theta)=1-sin^2(theta) . Put sin^2(th)=y so 5-5y=6y 11y=5 so y=5÷11 so sin(theta)=√5÷√11=.674

On applying square root on both sides, we get $\large \frac { \sin { \theta } }{\sqrt{ 5} } =\frac { \cos { \theta } }{ \sqrt{6 }} \\ \Rightarrow \tan{\theta} = \frac {\sqrt{5}}{ \sqrt{6}}$ By applying Pythagorean Identities $\sin { \theta } =\sqrt { \frac { 5 }{ 11 } } =0.674$ ( $\theta$ is acute angle so we have taken the positive value for $\sin{\theta}$ )