Question

If secp=5/4, then find sinP cosR+ cosP sinR. Given that triangle PQR is right angled at Q

Answer 1

Sec p=5/4=hyp/adj

By Pythogoras Theorem

PR^2=PQ^2+QR^2

(5)^2=(4)^2+QR^2

QR^2=5^2-4^2

QR^2=25-16

QR^2=9

QR=3CM

SinP=opp/hyp=3/5

CosP=adj/hyp=4/5

SinR=opp/hyp=4/5

CosR=adj/hyp=3/5

SinP CosR+CosP SinR

=3/5×3/5+4/5×4/5

=9/5+16/5

=25/5

=5

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