Question

In the given Fig, $\angle$R is the right angle of $∆$PQR. Write the following ratios.
(i) sin P (ii) cos Q (iii) tan P (iv) tan Q

Answer 1

(i) sinP = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{P}}{\mathrm{Hypotenuse}}=\frac{\mathrm{QR}}{\mathrm{PQ}}$

(ii) cosQ = $\frac{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{Q}}{\mathrm{Hypotenuse}}=\frac{\mathrm{QR}}{\mathrm{PQ}}$

(iii) tanP = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{P}}{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{P}}=\frac{\mathrm{QR}}{\mathrm{PR}}$

(iv) tanQ = $\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}\angle \mathrm{Q}}{\mathrm{Adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{Q}}=\frac{\mathrm{PR}}{\mathrm{QR}}$