Question

In fig 8.20, $∆$UVW is right angled triangle and $\angle$UVW = 90$°$, UV= 6cm, and UW = 8cm, find all trignometric ratios of angle W.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{UVW},\mathrm{we}\mathrm{have}: \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem},\mathrm{we}\mathrm{have}: \\ {\left(\mathrm{UW}\right)}^2={\left(\mathrm{UV}\right)}^2+{\left(\mathrm{VW}\right)}^2 \\ \Rightarrow 8^2=6^2+{\left(\mathrm{VW}\right)}^2 \\ \Rightarrow {\left(\mathrm{VW}\right)}^2=64-36=28 \\ \Rightarrow \mathrm{VW}=2\sqrt{7} \end{gathered}$$

Now, we have:
$$\begin{gathered} \sin \mathrm{W}=\frac{6}{8}=\frac{3}{4},\mathrm{cosec}\mathrm{W}=\frac{8}{6}=\frac{4}{3} \\ \cos \mathrm{W}=\frac{2\sqrt{7}}{8}=\frac{\sqrt{7}}{4},\sec \mathrm{W}=\frac{8}{2\sqrt{7}}=\frac{4}{\sqrt{7}} \\ \tan \mathrm{W}=\frac{6}{2\sqrt{7}}=\frac{3}{\sqrt{7}},\cot \mathrm{W}=\frac{2\sqrt{7}}{6}=\frac{\sqrt{7}}{3} \end{gathered}$$