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**RS Aggarwal Class 10 Solutions Chapter 5 Trigonometric Ratios**

RS Aggarwal Class 10 Solutions

**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

**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

Amresh says

Very best Solution