Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

9th Maths Trigonometry Exercise 6.1 Question 1.
From the given figure find all the trigonometric ratios of angle B.
Solution:

9th maths trigonometry exercise 6.1 Question 2.
From the given figure, find the values of
(i) sin B
(ii) sec B
(iii) cot B
(iv) cos C
(v) tan C
(vi) cosec C
Solution:

trigonometry exercise 6.1 Question 3.
If 2 cos θ = \(\sqrt{3}\), then find all the trigonometric ratios of angle θ.
Solution:

Trigonometry solutions exercise 6.1 Question 4.
If cos A = \(\frac{3}{5}\), then find the value of \(\frac{\sin A-\cos A}{2 \tan A}\)
Solution:

trignometry solution of 9th class Question 5.
If cos A = \(\frac{2 x}{1+x^{2}}\) then find the values of sin A and tan A in terms of x.
Solution:

9th Maths Exercise 6.1 Samacheer Kalvi Question 6.

Solution:

9th Maths Exercise 6.1 Question 7.
If 3 cot A = 2, then find the value of \(\frac{4 \sin A-3 \cos A}{2 \sin A+3 \cos A}\)
Solution:

9th Std Maths Trigonometry Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8 \cos \theta-2 \sin \theta}{4 \cos \theta+2 \sin \theta}\)
Solution:

9th Maths Exercise 6.1 Solution Question 9.
From the given figure, prove that θ + ϕ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ϕ
Solution:


(∴ By Pythagoras theorem, in a right angled triangle square of hypotenuse is equal to sum of the squares of other two side)
And also in the figure, ∆ADC, ∆DBC are two other triangles.
As per the data given,
9²+ 12² = 81 + 144 = 225 = 15²
∴ ∆ ADC is a right angled triangle, then 12² + 16² = 144 + 256 = 400 = 20²
∴ ∆ DBC is also a right angled triangle

9th Maths Trigonometry Question 10.
A boy standing at point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios)
Solution:
In the figure,
∆OPM, ∆OQN are similar triangles. In similar triangles the sides are in the same proportional.