How to identify sides?Identify the angle with respect to which the t-ratios have to be calculated. Sides are always labelled with respect to the ‘θ’ being considered.
Let us take two cases:
In a right triangle ABC, right-angled at B. Once we have identified the sides, we can define six t-Ratios with respect to the sides.
Note from above six relationships:
However, it is very tedious to write full forms of t-ratios, therefore the abbreviated notations are:
Example: If in a right-angled triangle ABC, right-angled at B, hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm and if ∠ACB = θ, then find tan θ, sin θ and cos θ.
Solution:
Given,
In ∆ABC,
Hypotenuse, AC = 5cm
Base, BC = 3cm
Perpendicular, AB = 4cm
Then,
tan θ = Perpendicular/Base = 4/3
Sin θ = Perpendicular/Hypotenuse = AB/AC = ⅘
Cos θ = Base/Hypotenuse = BC/AC = ⅗
An equation involving trigonometric ratio of angle(s) is called a trigonometric identity, if it is true for all values of the angles involved. These are:
Example: Express the ratios cosA, tanA, and secA in terms of sinA.
Solution: Since cos2A + sin2A = 1, therefore:
cos2A = 1 − sin2A
i.e., cosA = ±√1−sin2A
This gives:
cosA = √1 − sin2A
Hence,
and
Note: A trigonometric ratio only depends upon the angle ‘θ’ and stays the same for same angle of different sized right triangles.
Values of Trigonometric Ratios of Specified Angles
The value of sin θ and cos θ can never exceed 1 (one) as opposite side is 1. Adjacent side can never be greater than hypotenuse since hypotenuse is the longest side in a right-angled ∆.
Example: If tan θ + cot θ = 5, find the value of tan2θ + cotθ.
Solution:
tan θ + cot θ = 5 … [Given
tan2θ + cot2θ + 2 tan θ cot θ = 25 … [Squaring both sides
tan2θ + cot2θ + 2 = 25
∴ tan2θ + cot2θ = 23
Example: If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A.
Solution:
sec 2A = cosec (A – 27°)
cosec(90° – 2A) = cosec(A – 27°) …[∵ sec θ = cosec (90° – θ)
90° – 2A = A – 27°
90° + 27° = 2A + A
⇒ 3A = 117°
∴ ∠A = 117°/3 = 39°
123 videos|457 docs|77 tests
|
|
Explore Courses for Class 10 exam
|