Solution:

Given:

△PQR is right-angled at Q.

PQ = 4cm

RQ = 3cm

Required to find: sin P, sin R, sec P, sec R

Given △PQR,

By using Pythagoras theorem to △PQR, we get

PR² = PQ² +RQ²

Substituting the respective values,

PR² = 4² +3²

PR² = 16 + 9

PR² = 25

PR = √25

PR = 5

⇒ Hypotenuse =5

By definition,

sin P = Perpendicular side opposite to angle P/ Hypotenuse

sin P = RQ/ PR

⇒ sin P = 3/5

And,

sin R = Perpendicular side opposite to angle R/ Hypotenuse

sin R = PQ/ PR

⇒ sin R = 4/5

And,

sec P=1/cos P

secP = Hypotenuse/ Base side adjacent to ∠P

sec P = PR/ PQ

⇒ sec P = 5/4

Now,

sec R = 1/cos R

secR = Hypotenuse/ Base side adjacent to ∠R

sec R = PR/ RQ

⇒ sec R = 5/3