from a handpicked tutor in LIVE 1-to-1 classes
Given sec θ = 13/12, calculate all other trigonometric ratios
We will use the basic concept of trigonometric ratios to solve the problem.
Let ΔABC be a right-angled triangle, right angled at point B.
It is given that:
sec θ = hypotenuse / side adjacent to ∠θ = AC/AB = 13/12
Let AC = 13k and AB = 12k, where k is a positive integer.
Applying Pythagoras theorem in Δ ABC, we obtain:
AC2 = AB2 + BC2
BC2 = AC2 - AB2
BC2 = (13k)2 - (12k)2
BC2 = 169 k2 - 144 k2
BC2 = 25k2
BC = 5k
sin θ = side opposite to ∠θ / hypotenuse = BC/AC = 5/13
cos θ = side adjacent to ∠θ / hypotenuse = AB/AC = 12/13
tan θ = side opposite to ∠θ / side adjacent to ∠θ = BC/AB = 5/12
cot θ = side adjacent to ∠θ / side opposite to ∠θ = AB/BC = 12/5
cosec θ = hypotenuse / side opposite to ∠θ = AC/BC = 13/5
☛ Check: NCERT Solutions Class 10 Maths Chapter 8
Given sec θ = 13/12, calculate all other trigonometric ratios.
Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.1 Question 5
If sec θ = 13/12, all the other trigonometric ratios are as follows: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5 and cosec θ = 13/5 respectively.
☛ Related Questions:
- If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
- If cotθ = 7/8, evaluate: (i) (1 + sinθ)(1 - sinθ) / (1 + cosθ)(1 - cosθ), (ii) cot2θ
- If 3 cot A = 4, check whether (1 - tan2 A) / (1 + tan2 A) = cos2 A - sin2 A or not.
- In the triangle ABC right-angled at B, if tan A = 1/√3 find the value of:(i) sin A cos C + cos A sin C(ii) cos A cos C - sin A sin C