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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

Respuesta :

Answer:  (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

  • ∠R = 30°
  • ∠P = 90°
  • ∠Q = 60°

(i) sin² (90°) = 1² = 1

[tex]\text{(ii)}\ \cos ^2(30^o)= \bigg(\dfrac{\sqrt3}{2}\bigg)^2=\dfrac{3}{4}[/tex]

[tex].\quad \tan (30^o)=\dfrac{1}{\sqrt3}\bigg=\dfrac{\sqrt3}{3}[/tex]

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

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

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