# In the Fig. 8.12, ∠R is the right angle of ΔPQR. Write the following ratios.(i) sin P (ii) cos Q(iii) tan P (iv) tan Q For any right-angled triangle,

sinθ = Opposite side Side/Hypotenuse

tanθ = sinθ/cosθ

cotθ = 1/tanθ

secθ = 1/cosθ

cosecθ = 1/sinθ

= Hypotenuse/Opposite side Side

In the given triangle let us understand, the Opposite side and Adjacent sidesides.

So for P,

Opposite side Side = QR

So, for Q,

Opposite side Side = PR

In general for the side Opposite side to the 90° angle is the hypotenuse.

So, for Δ PQR, hypotenuse = PQ

(i) sin P = Opposite side Side/Hypotenuse

= QR/PQ

(ii) cos Q = Adjacent sideSide/Hypotenuse

= QR/PQ

(iii) tan P = sinθ/cosθ

= QR/PR

(iv) tan Q = sinθ/cosθ

= PR/QR

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