2)If triangle RPQ is a right angled triangle at Q. If PQ = 5cm and RQ = 10 cm, find (i) sin ^2 P (ii) cos ^2 R and tan R (iii) sin P X cos P (iV) sin ^2 P –cos ^2p

 (i) 1       (ii) 3/4 , √3/3       (iii) 0       (iv) 1

Step-by-step explanation:

ΔRPQ is 1 30°-60°-90° triangle because RQ = 2PQ , where

(i) sin² (90°) = 1² = 1

(iii) sin (90°) × cos (90°) = 1 × 0 = 0

(iv) sin² (90°) - cos² (90°) = 1² - 0² = 1

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345