In the given figure, ∠R is the right angle of ∆PQR. Write the foll

1.	In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.i. sin P ii. cos Q iii. tan P iv. tan Q
Answer» i. sin P = \(\cfrac{Opposite \,of ∠ P}{Hypotenuse}\) = \(\frac{QR}{PQ}\) ii .cos Q = \(\cfrac{Adjacent\, side \,of ∠ Q}{Hypotenuse}\) = \(\frac{QR}{PQ}\) iii. tan P = \(\cfrac{Opposite \,side\,of ∠ P}{Adjacent\, side \, of\,∠P}\) = \(\frac{QR}{PR}\) iv. tan Q = \(\cfrac{Opposite \,side\,of ∠ Q}{Adjacent\, side \, of\,∠Q}\) = \(\frac{PR}{QR}\)

In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.i. sin P ii. cos Q iii. tan P iv. tan Q

Answer»

i. sin P = \(\cfrac{Opposite \,of ∠ P}{Hypotenuse}\) = \(\frac{QR}{PQ}\)

ii .cos Q = \(\cfrac{Adjacent\, side \,of ∠ Q}{Hypotenuse}\) = \(\frac{QR}{PQ}\)

iii. tan P = \(\cfrac{Opposite \,side\,of ∠ P}{Adjacent\, side \, of\,∠P}\) = \(\frac{QR}{PR}\)

iv. tan Q = \(\cfrac{Opposite \,side\,of ∠ Q}{Adjacent\, side \, of\,∠Q}\) = \(\frac{PR}{QR}\)