## RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios

**Exercise 5**

**Question 1:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

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**Question 2:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

Let AB = 7k and AC = 25k,

Where k is positive

By Pythagoras theorem, we have

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**Question 3:**

Given:

Let AB = 15k and AC = 8k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

By Pythagoras theorem, we have

AC^{2 }= AB^{2 }+ BC^{2}

**Question 4:**

Given:

Let AB = 2k and AC = 1k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

By Pythagoras theorem, we have

AC^{2 }= AB^{2 }+ BC^{2}

**Question 5:**

Given:

Let AB = k and AC = 1k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

By Pythagoras theorem, we have

**Question 6:**

Given:

Let AB = 15k and BC = 8k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

By Pythagoras theorem, we have

AC^{2 }= AB^{2 }+ BC^{2}

**Question 7:**

Given:

Let BC = 4k and AB = 3k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

**Question 8:**

Given:

Let BC = 1k and AB = k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

**Question 9:**

Given:

Let AC = 2k and BC = 1k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

**Question 10:**

Given:

Let AC = 5k and AB = 4k,

Where k is positive

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

**Question 11:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠BAC = θ

**Question 12:**

Given: 3tanθ = 4

⇒

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠A = θ

**Question 13:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠A = θ

**Question 14:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠A = θ

By Pythagoras theorem, we have

**Question 15:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠A = θ

By Pythagoras theorem, we have

**Question 16:**

Given:

Let us draw a ∆ABC in which ∠B = 90^{0 }and ∠A = θ

By Pythagoras theorem, we have

AC^{2 }= AB^{2 }+ BC^{2}

**Question 17:**

Given: ∆ABC in which ∠B = 90^{0}, AB = 12cm, BC = 5cm

By Pythagoras theorem, we have

**Question 18:**

Given: ∆ABC in which ∠B = 90^{0}, AB = 24 cm and BC = 7cm

By Pythagoras theorem, we have

**Question 19:**

Given: ∆ABC in which ∠B = 90^{0 }and ∠A = θ, BC = 21cm units and AB = 29 units

By Pythagoras theorem, we have

Hope given RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

Sukanya says

Please give Q.33 of chap 5 Trigonometric Ratios of class 10

Tanya says

Please give me question 35 solution

shanu kumar says

please give me cce question solution please sir