If cot𝜽= 𝟒𝟎/𝟗, find the values of cosecθ and sinθ.

Chapter 6 – Trigonometry – Text Book Solution

Practice Set 6.1| Q 3 | Page 131

Solution
We know that $$cot(theta) = text{adjacent}/text{opposite}$$ in a right triangle, where $$theta$$ is one of the acute angles of the triangle.
Given $$cot(theta) = 40/9$$, we can label the adjacent side of the right triangle as 40 and the opposite side as 9.
Using the Pythagorean theorem, we can find the hypotenuse:
$$a^2 + b^2 = c^2$$
$$9^2 + 40^2 = c^2$$
$$81 + 1600 = c^2$$
$$1681 = c^2$$
$$c = sqrt{1681} = 41$$
Now we can use the definitions of the trigonometric functions to find the values of $$csc(theta)$$ and $$sin(theta)$$:
$$csc(theta) = text{hypotenuse}/text{opposite} = 41/9$$
$$sin(theta) = text{opposite}/text{hypotenuse} = 9/41$$
Therefore, $$csc(theta) = 41/9$$ and $$sin(theta) = 9/41$$.

Solution

We know that cot(θ) = adjacent/opposite in a right triangle, where θ is one of the acute angles of the triangle.

Given cot(θ) = 40/9, we can label the adjacent side of the right triangle as 40 and the opposite side as 9.

Using the Pythagorean theorem, we can find the hypotenuse:

a² + b² = c²

9² + 40² = c²

81 + 1600 = c²

1681 = c²

c = √1681 = 41

Now we can use the definitions of the trigonometric functions to find the values of cosec(θ) and sin(θ):

cosec(θ) = hypotenuse/opposite = 41/9

sin(θ) = opposite/hypotenuse = 9/41

Therefore, cosec(θ) = 41/9 and sin(θ) = 9/41.

Chapter 6 – Trigonometry – Text Book Solution

Practice set 6.1 |Q 2 | P 131

Q.2

Q 3

Q.4

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