Question

Solve Q No.11
Q.17. If cos x$°$ = sin 200$°$, find the possible values of x between - 180$°$ and 360$°$.

Answer 1

$$\begin{gathered} \cos \mathrm{x}=\sin 200° \\ \Rightarrow \sin \left(90°-\mathrm{x}\right)=\sin 200° \\ \Rightarrow 90°-\mathrm{x}=200° \\ \Rightarrow \mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{110}\mathbf{°} \\ \mathrm{Again},\cos \mathrm{x}=\sin 200° \\ \Rightarrow \cos \mathrm{x}=\cos \left(90°-200°\right) \\ \Rightarrow \cos \mathrm{x}=\cos \left(-110°\right) \\ \Rightarrow \cos \mathrm{x}=\cos 110°\left[\mathbf{as}\mathbf{,}\mathbf{}\mathbf{cos}\left(\mathbf{-}\mathbf{x}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\right] \\ \Rightarrow \mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{110}\mathbf{°} \\ \mathrm{Again}\cos \mathrm{x}=\sin 200° \\ \Rightarrow \cos \mathrm{x}=\cos \left(90°-200°\right) \\ \Rightarrow \cos \mathrm{x}=\cos \left(-110°\right) \\ \Rightarrow \cos \mathrm{x}=\cos 110° \\ \Rightarrow \cos \mathrm{x}=\cos \left(360°-110°\right) \\ \Rightarrow \cos \mathrm{x}=\cos 250° \\ \Rightarrow \mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{250}\mathbf{°} \end{gathered}$$