Trig Ratio Problems

Basic Trig ratio problems are very important when dealing with triangles. In the below questions we will learn how to find the values of the other ratio where one ratio is given.


1. If sec θ = 17/8 and θ is a positive acute angle, find the value of csc θ using Pythagoras theorem.

Solution:

Draw a right-angled ∆ ABC such that ∠ABC = θ,

Hypotenuse = BA = 17, and Adjacent side (or base) = BC = 8.

Then we get,

sec θ = 17/8



Now, from the right-angled ∆ ABC we get,

Trig Ratio Problems

AC2 + BC2 = BA2

⇒ AC2 = BA2 - BC2

⇒ AC2 = (17)2 - 82

⇒ AC2 = 289 - 64

⇒ AC2 = 225



Therefore, AC = 15 (Since θ is a positive acute angle so, AC is also positive)

Therefore, csc θ = BA/AC

⇒ csc θ = 17/15


In this question on Trig ratio problems we will learn how to find the value of sin θ when θ is a positive acute angle.


2. If tan θ + sec θ = 2/√3 and θ is a positive acute angle, find the value of sin θ.

Solution:

Given, tan θ + sec θ = 2/√3,

⇒ sin θ/cos θ + 1/cos θ  = 2/√3,( Since tan θ = sin θ/cos θ and sec θ = 1/cos θ)

⇒ (sin θ + 1)/cos θ = 2/√3            

⇒ √3 (sin θ + 1) = 2 cos θ

⇒ 3(sin θ + 1)2 = 4 cos2 θ, (Squaring both sides)

⇒ 3(sin2 θ + 2 sin θ + 1) = 4(1 - sin2 θ)

⇒ 3 sin2 θ + 6 sin θ + 3 = 4 - 4 sin2 θ

⇒ 3 sin2 θ + 6 sin θ + 3 - 4 + 4 sin2 θ = 0

⇒ 7 sin2 θ + 6 sin θ - 1 = 0

⇒ 7 sin2 θ + 7 sin θ - sin θ - 1 =0

⇒   7 sin θ (sin θ + 1) - 1 (sin θ + 1) =0         

⇒ (7 sin θ - 1)(sin θ + 1) = 0

Therefore,

Either, 7 sin θ - 1 = 0

⇒ 7 sin θ = 1

 ⇒ sin θ = 1/7


or, sin θ + 1 = 0

⇒ sin θ = - 1

According to the problem, θ is a positive acute angle; so, we neglect, sin θ = -1.

Therefore, sin θ = 1/7 

The below solved Trig ratio problems will help us to find the values of the ratio using trigonometric identity.


3. If θ is a positive acute angle and sec θ = 25/7, find the value of csc θ using trigonometric identity.

Solution:

Given, sec θ = 25/7

Therefore, cos θ = 1/sec θ

⇒ cos θ = 1/(25/7)

⇒ cos θ = 7/25

We know that, sin2 θ + cos2 θ = 1

⇒ sin2 θ = 1 - cos2 θ

⇒ sin2 θ = 1 - (7/25)2

⇒ sin2 θ = 1 - (49/625)

⇒ sin2 θ = (625 – 49)/625

⇒ sin2 θ = 576/625

Now, taking square root on both the sides we get,

⇒ sin θ  = 24/25 (Since θ is a positive acute angle so, sin θ is also positive)

Therefore, csc θ = 1/sin θ

⇒ csc θ = 1/(24/25)

⇒ csc θ = 25/24 . 

In this question on Trig ratio problems we will learn how to find the minimum value of the given T-ratio.

4. Find the minimum value of cos2 θ + sec2 θ

Solution:

cos2 θ + sec2 θ

= (cos θ)2 + (sec θ)2 - 2 cos θ ∙ sec θ + 2 cos θ sec θ

= (cos θ - sec θ)2 + 2 ∙ 1 (since, cos θ ∙ sec θ = 1)

= (cos θ - sec θ)2 + 2

(cos θ - sec θ)2 ≥ 0

Therefore, (cos θ - sec θ)2 + 2 ≥ 2 (since, adding 2 on both the sides)

i.e., cos2 θ + sec2 θ ≥ 2

Therefore, the minimum value of cos2 θ + sec2 θ is 2.

 Trigonometric Functions





10th Grade Math

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