Q. 44.1( 14 Votes )

# In the figure 8.1

Answer :

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

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

So, for Δ PQR,

So, for α,

Opposite side Side = PQ

Adjacent sideSide = QR

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

So, for Δ PQR, hypotenuse = PR

sin α = Opposite side Side/Hypotenuse

= PQ/PR

cos α = Adjacent sideSide/Hypotenuse

= QR/PR

tan α = sinθ/cosθ

= Opposite side Side/Adjacent sideSide

= PQ/QR

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

So for Δ PQS,

So for θ,

Opposite side Side = QS

Adjacent sideSide = PQ

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

So for Δ PQS, hypotenuse = PS

sinθ = Opposite side Side/Hypotenuse

= QS/PS

cosθ = Adjacent sideSide/Hypotenuse

= PQ/PS

tanθ = sinθ/cosθ

= Opposite side Side/Adjacent sideSide

= QS/PQ

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