(200) to the power – 10 + x (20) to the power 10 x (2) to the power 10 = (8) to the power 50​

1 thought on “(200) to the power – 10 + x (20) to the power 10 x (2) to the power 10 = (8) to the power 50​”

1. 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.

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