Question

if sin theta= 3/5 and if theta lies in 3rd quadrant find the value of all t-ratios​

Answer 1

Step-by-step explanation:

please see the above image for better understanding

Answer 2

Given :-

${sin\theta} = \dfrac{3}{5}$

To find :-

All Trigonometric ratios in 3rd Quadrant

SOLUTION:-

In a third Quadrant

• sinθ is negative
• cosθ is negative
• tanθ is negative
• cscθ is negative
• secθ is negative
• cotθ is positive

So,

${sin\theta} = \dfrac{3}{5}$

As we know that ,

${sin\theta} = \dfrac{opposite side}{Hypotenuse}$

So,

• opposite side = 3
• Hypotenuse = 5

From , Pythagoras theorem we can find adjacent side

(opp)² + (adj)² = (hyp)²

(3)² + (adj)² = (5)²

9 + (adj)² = 25

(adj)² = 16

adjacent = 4

So, Now All we know that

${sin\theta} = \dfrac{Opposite}{Hypotenuse}$

${cos\theta} = \dfrac{Adjacent}{Hypotenuse}$

${tan\theta} = \dfrac{Opposite}{Adjacent}$

${csc\theta} = \dfrac{Hypotenuse}{Opposite}$

${sec\theta} = \dfrac{Hypotenuse}{Adjacent}$

${cot\theta} = \dfrac{Adjacent}{opposite}$

But in 3rd Quadrant ,

sinθ, cosθ , cscθ , secθ are negative

So, we get

${sin\theta} = \dfrac{-Opposite}{Hypotenuse}$

☆${sin\theta} = \dfrac{-3}{5}$

${cos\theta} = \dfrac{-Adjacent}{Hypotenuse}$

☆ ${cos\theta} = \dfrac{-4}{5}$

${csc\theta} = \dfrac{-Hypotenuse}{Opposite}$

☆${csc\theta} = \dfrac{-5}{3}$

${sec\theta} = \dfrac{-Hypotenuse}{Adjacent}$

☆ ${sec\theta} = \dfrac{-5}{4}$

As tanθ and cotθ are positive in 3rd Quadrant so,

${tan\theta} = \dfrac{Opposite}{Adjacent}$

☆ ${tan\theta} = \dfrac{3}{4}$

${cot\theta} = \dfrac{Adjacent}{opposite}$

☆ ${cot\theta} = \dfrac{4}{3}$