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

cosθ = Adjacent sideSide/Hypotenuse

tanθ = sinθ/cosθ

= Opposite side Side/Adjacent sideSide

cotθ = 1/tanθ

= Adjacent sideSide/Opposite side Side

secθ = 1/cosθ

= Hypotenuse/Adjacent sideSide

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

Adjacent sideSide = PR

So, for ∠ Q,

Opposite side Side = PR

Adjacent sideSide = QR

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

= Opposite side Side/Adjacent sideSide

= QR/PR

(iv) tan Q = sinθ/cosθ

= Opposite side Side/Adjacent sideSide

= PR/QR