Question

7.
In a APQR, right angled at Q, PQ = 15 cm, QR = 8 cm. Determine:
sin P
(b) cos P
(a)​

Answer 1

Actually the question should be: (a) Sin Q

(b)Cos Q

Ans:-If triangle PQR is a right angled triangle then,

Perpendicular(PQ)^2+Base(QR)^2=Hypotenuse(PR)^2 [Hypotenuse=x]

(15)^2+(8)^2=x^2

225+64=289

x=17

(a)cos Q=Base/Hypotenuse

= 8/17

(b)sin Q=Perpendicular/Hypotenuse

= 15/17