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

RS Aggarwal Class 10 Solutions

**Exercise 5**

**Question 1:**

Given: [latex s=2]sin\theta =\frac { \sqrt { 3 } }{ 2 }[/latex]

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

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**Question 2:**

Given: [latex s=2]cos\theta =\frac { \sqrt { 7 } }{ 25 }[/latex]

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: [latex s=2]tan\theta =\frac { BC }{ AB } =\frac { 15 }{ 8 } [/latex]

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: [latex s=2]cot\theta =\frac { AB }{ BC } =\frac { 2 }{ 1 } [/latex]

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: [latex s=2]cosec\theta =\frac { AC }{ BC } =\frac { \sqrt { 10 } }{ 1 } [/latex]

Let AB = [latex s=2]\sqrt { 10 } [/latex]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: [latex s=2]cot\theta =\frac { AB }{ BC } =\frac { 15 }{ 8 } [/latex]

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: [latex s=2]tan\theta =\frac { BC }{ AB } =\frac { 4 }{ 3 } [/latex]

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: [latex s=2]tan\theta =\frac { BC }{ AB } =\frac { 1 }{ \sqrt { 7 } } [/latex]

Let BC = 1k and AB = [latex s=2]\sqrt { 7 } [/latex]k,

Where k is positive

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

**Question 9:**

Given: [latex s=2]cosec\theta =\frac { AC }{ BC } =\frac { 2 }{ 1 } [/latex]

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: [latex s=2]sec\theta =\frac { AC }{ AB } =\frac { 5 }{ 4 } [/latex]

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: [latex s=2]tan\theta =0.6=\frac { 6 }{ 10 } =\frac { 3 }{ 5 } [/latex]

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

**Question 12:**

Given: 3tanθ = 4

⇒ [latex s=2]tan\theta =\frac { 4 }{ 3 } [/latex]

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

**Question 13:**

Given: [latex s=2]tan\theta =\frac { a }{ b } =\frac { ak }{ bk } =\frac { BC }{ AB } [/latex]

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

**Question 14:**

Given: [latex s=2]cot\theta =\frac { 2 }{ 3 } =\frac { 2k }{ 3k } [/latex]

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

By Pythagoras theorem, we have

**Question 15:**

Given: [latex s=2]sec\theta =\frac { 17 }{ 8 } =\frac { 17k }{ 8k } [/latex]

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

By Pythagoras theorem, we have

**Question 16:**

Given: [latex s=2]tan\theta =\frac { 20 }{ 21 } =\frac { 20k }{ 21k } [/latex]

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