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Question

# In the given figure, $\angle$PQR = 90° , $\angle$PQS = 90° , $\angle$PRQ = $\mathrm{\alpha }$ and$\angle$QPS = $\mathrm{\theta }$ Write the following trigonometric ratios. (i) sin$\mathrm{\alpha }$, cos$\mathrm{\alpha }$ , tan$\mathrm{\alpha }$ (ii) sin$\mathrm{\theta }$, cos$\mathrm{\theta }$, tan$\mathrm{\theta }$

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Solution

## (i) sin$\alpha$ = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{PRQ}}{\mathrm{Hypotenuse}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$ cos$\alpha$ = $\frac{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{PRQ}}{\mathrm{Hypotenuse}}=\frac{\mathrm{RQ}}{\mathrm{PR}}$ tan$\alpha$ = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{PRQ}}{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{PRQ}}=\frac{\mathrm{PQ}}{\mathrm{RQ}}$ (ii) sinθ = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{QPS}}{\mathrm{Hypotenuse}}=\frac{\mathrm{QS}}{\mathrm{PS}}$ cosθ = $\frac{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{QPS}}{\mathrm{Hypotenuse}}=\frac{\mathrm{PQ}}{\mathrm{PS}}$ tanθ = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{QPS}}{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{QPS}}=\frac{\mathrm{QS}}{\mathrm{PQ}}$

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