Question

In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

sin $\mathrm{\theta }$		$\frac{11}{61}$		$\frac{1}{2}$				$\frac{3}{5}$
cos $\mathrm{\theta }$	$\frac{35}{37}$				$\frac{1}{\sqrt{3}}$
tan $\mathrm{\theta }$			1			$\frac{21}{20}$	$\frac{8}{15}$		$\frac{1}{2\sqrt{2}}$

Answer 1

(i)
$\cos \theta =\frac{35}{37}$

Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\sin }^2\theta +{\left(\frac{35}{37}\right)}^2=1 \\ \Rightarrow {\sin }^2\theta =1-\frac{1225}{1369}=\frac{1369-1225}{1369}=\frac{144}{1369} \\ \Rightarrow {\sin }^2\theta ={\left(\frac{12}{37}\right)}^2 \\ \Rightarrow \sin \theta =\frac{12}{37} \end{gathered}$$
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \frac{12}{37}}}{{\displaystyle \frac{35}{37}}}=\frac{12}{35}$

(ii)
$\sin \theta =\frac{11}{61}$

Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(\frac{11}{61}\right)}^2+{\cos }^2\theta =1 \\ \Rightarrow {\cos }^2\theta =1-\frac{121}{3721}=\frac{3721-121}{3721}=\frac{3600}{3721} \\ \Rightarrow {\cos }^2\theta ={\left(\frac{60}{61}\right)}^2 \\ \Rightarrow \cos \theta =\frac{60}{61} \end{gathered}$$
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \frac{11}{61}}}{{\displaystyle \frac{60}{61}}}=\frac{11}{60}$

(iii)
$$\begin{gathered} \tan \theta =1 \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{1}{1} \\ \Rightarrow \sin \theta =k,\cos \theta =k \end{gathered}$$
Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow k^2+k^2=1 \\ \Rightarrow 2k^2=1 \\ \Rightarrow k^2=\frac{1}{2} \\ \Rightarrow k=\frac{1}{\sqrt{2}} \end{gathered}$$
$\therefore \sin \theta =k=\frac{1}{\sqrt{2}}$ and $\cos \theta =k=\frac{1}{\sqrt{2}}$

(iv)
$\sin \theta =\frac{1}{2}$

Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(\frac{1}{2}\right)}^2+{\cos }^2\theta =1 \\ \Rightarrow {\cos }^2\theta =1-\frac{1}{4}=\frac{4-1}{4}=\frac{3}{4} \\ \Rightarrow {\cos }^2\theta ={\left(\frac{\sqrt{3}}{2}\right)}^2 \\ \Rightarrow \cos \theta =\frac{\sqrt{3}}{2} \end{gathered}$$
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}=\frac{1}{\sqrt{3}}$

(v)
$\cos \theta =\frac{1}{\sqrt{3}}$

Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\sin }^2\theta +{\left(\frac{1}{\sqrt{3}}\right)}^2=1 \\ \Rightarrow {\sin }^2\theta =1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3} \\ \Rightarrow {\sin }^2\theta ={\left(\sqrt{\frac{2}{3}}\right)}^2 \\ \Rightarrow \sin \theta =\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}} \end{gathered}$$
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}}{{\displaystyle \frac{1}{\sqrt{3}}}}=\sqrt{2}$

(vi)
$$\begin{gathered} \tan \theta =\frac{21}{20} \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{21}{20} \\ \Rightarrow \sin \theta =21k,\cos \theta =20k \end{gathered}$$
Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(21k\right)}^2+{\left(20k\right)}^2=1 \\ \Rightarrow 441k^2+400k^2=1 \\ \Rightarrow 841k^2=1 \\ \Rightarrow k^2=\frac{1}{841}={\left(\frac{1}{29}\right)}^2 \\ \Rightarrow k=\sqrt{{\left(\frac{1}{29}\right)}^2}=\frac{1}{29} \end{gathered}$$
$\therefore \sin \theta =21k=21\times \frac{1}{29}=\frac{21}{29}$ and $\cos \theta =20k=20\times \frac{1}{29}=\frac{20}{29}$

(vii)
$$\begin{gathered} \tan \theta =\frac{8}{15} \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{8}{15} \\ \Rightarrow \sin \theta =8k,\cos \theta =15k \end{gathered}$$
Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(8k\right)}^2+{\left(15k\right)}^2=1 \\ \Rightarrow 64k^2+225k^2=1 \\ \Rightarrow 289k^2=1 \\ \Rightarrow k^2=\frac{1}{289}={\left(\frac{1}{17}\right)}^2 \\ \Rightarrow k=\sqrt{{\left(\frac{1}{17}\right)}^2}=\frac{1}{17} \end{gathered}$$
$\therefore \sin \theta =8k=8\times \frac{1}{17}=\frac{8}{17}$ and $\cos \theta =15k=15\times \frac{1}{17}=\frac{15}{17}$

(viii)
$\sin \theta =\frac{3}{5}$

Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(\frac{3}{5}\right)}^2+{\cos }^2\theta =1 \\ \Rightarrow {\cos }^2\theta =1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25} \\ \Rightarrow {\cos }^2\theta ={\left(\frac{4}{5}\right)}^2 \\ \Rightarrow \cos \theta =\frac{4}{5} \end{gathered}$$
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}=\frac{3}{4}$

(ix)
$$\begin{gathered} \tan \theta =\frac{1}{2\sqrt{2}} \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{1}{2\sqrt{2}} \\ \Rightarrow \sin \theta =k,\cos \theta =2\sqrt{2}k \end{gathered}$$
Now,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow k^2+{\left(2\sqrt{2}k\right)}^2=1 \\ \Rightarrow k^2+8k^2=1 \\ \Rightarrow 9k^2=1 \\ \Rightarrow k^2=\frac{1}{9}={\left(\frac{1}{3}\right)}^2 \\ \Rightarrow k=\sqrt{{\left(\frac{1}{3}\right)}^2}=\frac{1}{3} \end{gathered}$$
$\therefore \sin \theta =k=\frac{1}{3}$ and $\cos \theta =2\sqrt{2}k=2\sqrt{2}\times \frac{1}{3}=\frac{2\sqrt{2}}{3}$

The complete table is given below:

sinθ	$\frac{12}{37}$	$\frac{11}{61}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$\frac{\sqrt{2}}{\sqrt{3}}$	$\frac{21}{29}$	$\frac{8}{17}$	$\frac{3}{5}$	$\frac{1}{3}$
cosθ	$\frac{35}{37}$	$\frac{60}{61}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$	$\frac{20}{29}$	$\frac{15}{17}$	$\frac{4}{5}$	$\frac{2\sqrt{2}}{3}$
tanθ	$\frac{12}{35}$	$\frac{11}{60}$	1	$\frac{1}{\sqrt{3}}$	$\sqrt{2}$	$\frac{21}{20}$	$\frac{8}{15}$	$\frac{3}{4}$	$\frac{1}{2\sqrt{2}}$