In a right-angled triangle, the ratio of any two sides is known as the Trigonometric Ratios of Acute angles. Each trigonometric function has a different ratio.
Base (BC) – Adjacent side of ∠C
Perpendicular (AB) – opposite side of ∠C
Hypotenuse (AC) – opposite side of ∠B(right angle)
In the above right-angled triangle, ∠B is a right angle and there are two acute angles ∠A and ∠C. for both acute angles trigonometric ratios can be found.
Trigonometric Ratios of Acute Angle ∠C
Let ∠C = ϴ
1) sin ϴ = Perpendicular/Hypotenuse = AB/AC
2) cos ϴ = Base/Hypotenuse = BC/AC
3) tan ϴ = Perpendicular/Base = AB/BC
4) cosec ϴ = Hypotenuse/Perpendicular = AC/AB
5) sec ϴ = Hypotenuse/Base = AC/BC
6) cot ϴ = Base/Perpendicular = BC/AB
We can also find all the trigonometric ratios for the acute angle A. for acute angle A Base (Adjacent side of ∠A) will be AB and the Perpendicular (opposite side of ∠A) will be BC. hypotenuse will remain the same (AC).
Note – 1) In Trigonometry, trigonometric ratios depend on the acute angle ϴ, not on the size of the Right-angled triangle.
2) The trigonometric ratios are unique for each acute angle ϴ.
3) sin ϴ ≠ sin⨯ϴ , cos ϴ ≠ cos⨯ϴ, tan ϴ ≠ tan⨯ϴ
It is not multiplication between trigonometric function and acute angle.
4) Trigonometric ratios of any positive acute angle are always positive.
Some Examples –
Example – 1) In △ABC, ∠C is a right angle. side AB = 5 cm and BC = 4 cm then find all trigonometric ratios of angle A.
∵ △ABC is a right-angled triangle.
By Pythagoras theorem,
AC2 = AB2 – BC2
AC = √(5)2 – (4)2
AC = √(25-16) = √9 = 3 cm
|sin A = BC/AB = 4/5||cosec A = AB/BC = 5/4 (Reciprocal of sin A)|
|cos A = AC/AB = 3/5||sec A = AB/AC = 5/3 (Reciprocal of cos A)|
|tan A = BC/AC = 4/3||cot A = AC/BC = 3/4 (Reciprocal of tan A)|
Example – 2) if sin ϴ = 24/25, then find the remaining trigonometric ratios of ϴ.
Solution – sin ϴ = 24/25 = Perpendicular/Hypotenuse
If we draw △ABC for sin ϴ = 24/25
Perpendicular = AB and Hypotenuse = AC
Let AB = 24k, AC = 25k
Where k = proportionality constant
In △ABC By Pythagoras theorem,
BC = √(AC)2 – (AB)2
BC = √(25k)2 – (24k)2
BC = √625k2 – 576k2
BC = √49k2
BC = ±7k
since angle ϴ is an acute angle so, BC will be positive.
Now cos ϴ = 7k/25k = 7/25
tan ϴ = 24k/7k = 24/7
cosec ϴ = 25k/24k = 25/24
sec ϴ = 25k/7k = 25/7
cot ϴ = 7k/24k = 7/24 Ans.
Trigonometric Ratios of Acute Angles Class 10th in Hindi
2 thoughts on “Trigonometric Ratios of Acute Angles Class 10th”
Mү brother ѕuggеsteԁ I mіgһt likе this website. Ꮋe was totally right.
This post tｒuly made my day. You cann’t imagine
sіmply how muсh time I had spent for this information! Thanks!
Thanks for your compliment.