Question

If $\sin \theta =\frac{12}{13}$ then evaluate $\left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)$.

Answer 1

$$\begin{gathered} \mathrm{Given}:\sin \theta =\frac{12}{13} \\ \mathrm{Since},\sin \theta =\frac{P}{H} \\ \Rightarrow P=12\mathrm{and}H=13 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow {12}^2+B^2={13}^2 \\ \Rightarrow B^2=169-144 \\ \Rightarrow B^2=25 \\ \Rightarrow B=5 \\ \mathrm{Therefore}, \\ \cos \theta =\frac{B}{H}=\frac{5}{13} \\ \mathrm{Now}, \\ \left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)=\left(\frac{2\left({\displaystyle \frac{12}{13}}\right)-3\left({\displaystyle \frac{5}{13}}\right)}{4\left({\displaystyle \frac{12}{13}}\right)-9\left({\displaystyle \frac{5}{13}}\right)}\right) \\ =\left(\frac{{\displaystyle \frac{24}{13}}-{\displaystyle \frac{15}{13}}}{{\displaystyle \frac{48}{13}}-{\displaystyle \frac{45}{13}}}\right) \\ =\left(\frac{{\displaystyle 24-15}}{{\displaystyle 48-45}}\right) \\ =\frac{{\displaystyle 9}}{{\displaystyle 3}} \\ =3 \\ \mathrm{Hence},\left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)=3. \end{gathered}$$