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°

Basic Trigonometric Ratios 

Relations Between the Trigonometric Ratios

Problems on Trigonometric Ratios

Reciprocal Relations of Trigonometric Ratios

Trigonometrical Identity

Problems on Trigonometric Identities

Elimination of Trigonometric Ratios 

Eliminate Theta between the equations

Problems on Eliminate Theta 

Trig Ratio Problems

Proving Trigonometric Ratios

Trig Ratios Proving Problems

Verify Trigonometric Identities 





10th Grade Math

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