Question

in triangle pqr angle r =90,pq=29,qr=21,angle pqr=tita.find cos squaretita-sin2tita.​

Answer 1

Answer:

here in question a small mistake is happened . ∆PQR is right angled at R not at P if PQ = 29{hypotenuse}

a triangle PQR, right angled at R, in which PQ = 29 units, QR = 21 units and ∠PQR = θ.

so, PR = √{29² - 21²}

= √{(29-21)(29+21)}

= √{8 × 50}

= 20 unit

(i) cos²θ + sin²θ

from right angled triangle PQR ,

cosθ = QR/PQ = 21/29

sinθ = PR/PQ = 20/29

now, sin²θ = 400/841

cos²θ = 441/841

so, sin²θ + cos²θ = 400/841 + 441/841

= (400 + 441)/841 = 841/841 = 1

(ii) cos²θ - sin²θ

= (21/29)² - (20/29)²

= 441/841 - 400/841

= (441 - 400)/841

= 41/841

Step-by-step explanation:

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