Answer:

Some trigonometric solutions based problems on trigonometric ratios are shown here with the step-by-step explanation.

1. If sin θ = 8/17, find other trigonometric ratios of <θ.

Solution:

Problems on Trigonometric Ratios

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Let US DRAW a ∆ OMP in which ∠M = 90°.

Then sin θ = MP/OP = 8/17.

Let MP = 8k and OP = 17K, where k is positive.

By PYTHAGORAS’ theorem, we get

OP2 = OM2 + MP2

⇒ OM2 = OP2 – MP2

⇒ OM2 = [(17k)2 – (8k)2]

⇒ OM2 = [289k2 – 64k2]

⇒ OM2 = 225k2

⇒ OM = √(225k2)

⇒ OM = 15k

Therefore, sin θ = MP/OP = 8k/17k = 8/17

cos θ = OM/OP = 15k/17k = 15/17

tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15

csc θ = 1/sin θ = 17/8

sec θ = 1/cos θ = 17/15 and

cot θ = 1/tan θ = 15/8.

Step-by-step explanation:

Some trigonometric solutions based problems on trigonometric ratios are shown here with the step-by-step explanation.

1. If sin θ = 8/17, find other trigonometric ratios of <θ.

Solution:

Problems on Trigonometric Ratios

0Save

Let us draw a ∆ OMP in which ∠M = 90°.

Then sin θ = MP/OP = 8/17.

Let MP = 8k and OP = 17k, where k is positive.

By Pythagoras’ theorem, we get

OP2 = OM2 + MP2

⇒ OM2 = OP2 – MP2

⇒ OM2 = [(17k)2 – (8k)2]

⇒ OM2 = [289k2 – 64k2]

⇒ OM2 = 225k2

⇒ OM = √(225k2)

⇒ OM = 15k

Therefore, sin θ = MP/OP = 8k/17k = 8/17

cos θ = OM/OP = 15k/17k = 15/17

tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15

csc θ = 1/sin θ = 17/8

sec θ = 1/cos θ = 17/15 and

cot θ = 1/tan θ = 15/8.