Problems on Trigonometric Ratios

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



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.


2. If Cos A = 9/41, find other trigonometric ratios of ∠A.

Solution:

Problems on Trigonometric Ratio


Let us draw a ∆ ABC in which ∠B = 90°.

Then cos θ = AB/AC = 9/41.

Let AB = 9k and AC = 41k, where k is positive.





By Pythagoras’ theorem, we get

AC2 = AB2 + BC2

⇒ BC2 = AC2 – AB2

⇒ BC2 = [(41k)2 – (9k)2]

⇒ BC2 = [1681k2 – 81k2]

⇒ BC2 = 1600k2

⇒ BC = √(1600k2)

⇒ BC = 40k

Therefore, sin A = BC/AC = 40k/41k = 40/41

cos A = AB/AC = = 9k/41k = 9/41

tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9

csc A = 1/sin A = 41/40

sec A = 1/cos A = 41/9 and

cot A = 1/tan A = 9/40.


3. Show that the value of sin θ and cos θ cannot be more than 1.

Solution:

We know, in a right angle triangle the hypotenuse is the longest side.

Examples on Trigonometric Ratios










sin θ = perpendicular/hypotenuse = MP/OP < 1 since perpendicular cannot be greater than hypotenuse; sin θ cannot be more than 1.

Similarly, cos θ = base/hypotenuse = OM/OP < 1 since base cannot be greater than hypotenuse; cos θ cannot be more than 1.

4. Is that possible when A and B be acute angles, sin A = 0.3 and cos B = 0.7?

Solution:

Since A and B are acute angles, 0 ≤ sin A ≤ 1 and 0 ≤ cos B ≤ 1, that means the value of sin A and cos B lies between 0 to 1. So, it is possible that sin A = 0.3 and cos B = 0.7


5. If 0° ≤ A ≤ 90° can sin A = 0.4 and cos A = 0.5 be possible?

Solution:

We know that sin2A + cos2A = 1

Now put the value of sin A and cos A in the above equation we get;

(0.4)2 + (0.5)2 = 0.41 which is ≠ 1, sin A = 0.4 and cos A = 0.5 cannot be possible.


6. If sin θ = 1/2, show that (3cos θ - 4 cos3 θ) =0.

Solution:

Example on Trigonometric Ratios

Let us draw a ∆ ABC in which ∠B = 90° and ∠BAC = θ.

Then sin θ = BC/AC = 1/2.

Let BC = k and AC = 2k, where k is positive.





By Pythagoras’ theorem, we get

AC2 = AB2 + BC2

⇒ AB2 = AC2 – BC2

⇒ AB2 = [(2k)2 – k2]

⇒ AB2 = [4k2 – k2]

⇒ AB2 = 3k2

⇒ AB = √(3k2)

⇒ AB = √3k.

Therefore, cos θ = AB/AC = √3k/2k = √3/2

Now, (3cos θ - 4 cos3 θ)

= 3√3/2 - 4 ×(√3/2)3

= 3√3/2 - 4 × 3√3/8

= 3√3/2 - 3√3/2

= 0

Hence, (3cos θ - 4 cos<sup>3</sup> θ) = 0.


7. Show that sin α + cos α > 1 when 0° ≤ α ≤ 90°

Solution:

Trigonometric Problems











From the right triangle MOP,

Sin α = perpendicular/ hypotenuse

Cos α = base/ hypotenuse

Now, Sin α + Cos α

= perpendicular/ hypotenuse + base/ hypotenuse

= (perpendicular + base)/hypotenuse, which is > 1, Since we know that the sum of two sides of a triangle is always greater than the third side.


8. If cos θ = 3/5, find the value of (5csc θ - 4 tan θ)/(sec θ + cot θ)

Solution:

Trigonometric Problem

Let us draw a ∆ ABC in which ∠B = 90°.

Let ∠A = θ°

Then cos θ = AB/AC = 3/5.

Let AB = 3k and AC = 5k, where k is positive.

By Pythagoras’ theorem, we get

AC2 = AB2 + BC2

⇒ BC2 = AC2 – AB2

⇒ BC2 = [(5k)2 – (3k)2]

⇒ BC2 = [25k2 – 9k2]

⇒ BC2 = 16k2

⇒ BC = √(16k2)

⇒ BC = 4k

Therefore, sec θ = 1/cos θ = 5/3

tan θ = BC/AB =4k/3k = 4/3

cot θ = 1/tan θ = 3/4 and

csc θ = AC/BC = 5k/4k = 5/4

Now (5csc θ -4 tan θ)/(sec θ + cot θ)

= (5 × 5/4 - 4 × 4/3)/(5/3 + 3/4)

= (25/4 -16/3)/(5/3 +3/4)

= 11/12 × 12/29

= 11/29


9. Express 1 + 2 sin A cos A as a perfect square.

Solution:

1 + 2 sin A cos A

= sin2 A + cos2 A + 2sin A cos A, [Since we know that sin2 θ + cos2 θ = 1]

= (sin A + cos A)2


10. If sin A + cos A = 7/5 and sin A cos A =12/25, find the values of sin A and cos A.

Solution:

sin A + cos A = 7/5

⇒ cos A = 7/5 - sin θ

Now from sin θ/cos θ = 12/25

We get, sin θ(7/5 - sin θ) = 12/25

or, 7 sin θ – 5 sin2 θ = 12/5

or, 35 sin θ - 35 sin2 θ = 12

or, 25sin2 θ -35 sin θ + 12 = 0

or, 25 sin2 θ -20 sin θ - 15 sin θ + 12 = 0

or, 5 sin θ(5 sin θ - 4) - 3(5 sin θ - 4) = 0

or, (5 sin θ - 3) (5 sin θ - 4) = 0

⇒ (5 sin θ - 3) = 0 or, (5 sin θ - 4) = 0

⇒ sin θ = 3/5 or, sin θ = 4/5

When sin θ = 3/5, cos θ = 12/25 × 5/3 = 4/5

Again, when sin θ = 4/5, cos θ = 12/25 × 5/4 = 3/5

Therefore, sin θ =3/5, cos θ = 4/5

or, sin θ =4/5, cos θ = 3/5.


11. If 3 tan θ = 4, evaluate (3sin θ + 2 cos θ)/(3sin θ - 2cos θ).

Solution: Given,

3 tan θ = 4

⇒ tan θ = 4/3

Now,

(3sin θ + 2 cos θ)/(3sin θ - 2cos θ)

= (3 tan θ + 2)/(3 tan θ - 2), [dividing both numerator and denominator by cos θ]

= (3 × 4/3 + 2)/(3 × 4/3 -2), putting the value of tan θ = 4/3

= 6/2

= 3.


12. If (sec θ + tan θ)/(sec θ - tan θ) = 209/79, find the value of θ.

Solution: (sec θ + tan θ)/(sec θ - tan θ) = 209/79

⇒ [(sec θ + tan θ) - (sec θ - tan θ)]/[(sec θ + tan θ) + (sec θ - tan θ)] = [209 – 79]/[209 + 79], (Applying componendo and dividendo)

⇒ 2 tan θ/2 sec θ =130/288

⇒ sin θ/cos θ × cos θ = 65/144

⇒ sin θ = 65/144.


13. If 5 cot θ = 3, find the value of (5 sin θ - 3 cos θ)/(4 sin θ + 3 cos θ).

Solution:

Given 5 cot θ = 3

⇒ cot θ = 3/5

Now (5 sin θ - 3 cos θ)/(4 sin θ + 3 cos θ)

= (5 - 3 cot θ)/(4 sin θ + 3 cot θ), [dividing both numerator and denominator by sin θ]

= (5 - 3 × 3/5)/(4 + 3 × 3/5)

= (5 - 9/5)/(4 + 9/5)

= (16/5 × 5/29)

= 16/29.


13. Find the value of θ (0° ≤ θ ≤ 90°), when sin2 θ - 3 sin θ + 2 = 0

Solution:

⇒ sin2 θ -3 sin θ + 2 = 0

⇒ sin2 θ – 2 sin θ – sin θ + 2 = 0

⇒ sin θ(sin θ - 2) - 1(sin θ - 2) = 0

⇒ (sin θ - 2)(sin θ - 1) = 0

⇒ (sin θ - 2) = 0 or, (sin θ - 1) = 0

⇒ sin θ = 2 or, sin θ = 1

So, value of sin θ can’t be greater than 1,

Therefore sin θ = 1

⇒ θ = 90°

10th Grade Math

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