RBSE Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Exercise 14.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Trigonometric Ratios of Acute Angles |

Exercise |
Ex 14.1 |

Number of Questions Solved |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.1

Question 1.

If in a triangle ABC, ∠A = 90°, a = 25 cm, b = 7 cm, then find all (RBSESolutions.com) the trigonometrical ratios of ∠B and ∠C.

Solution.

Here it is given that

∠A = 90°, BC = 25 cm and AC = 7 cm

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

⇒ AB^{2} = BC^{2} – AC^{2}

Question 2.

If in ∆ABC, ∠B = 90°, a = 12 cm, b = 13 cm, then find (RBSESolutions.com) the trigonometrical ratios of ∠A and ∠C.

Solution.

Given that ∠B = 90°, AC = 13 cm, BC = 12 cm

Question 3.

If tan A = √2 – 1 then prove that sin A cos A = \(\frac { 1 }{ 2\surd 2 }\)

Solution.

tan A = \(\frac { \surd 2-1 }{ 1 } =\frac { Perpendicular }{ Base }\)

Now construct (RBSESolutions.com) a triangle ABC in which

∠B = 90° and BC : AB = (√2 – 1) : 1

⇒ BC = (√2 – 1 )k, AB = k, where k is any constant

Question 4.

If sin A = \(\frac { 1 }{ 3 }\) then find (RBSESolutions.com) the value of cos A . cosec A + tan A . sec A

Solution.

sin A = \(\frac { 1 }{ 3 }\)

Now we will construct a triangle ABC in which ∠B = 90° and BC : AC = 1 : 3

Let BC = k and AC = 3k where k > 0 which is a constant quantity.

Question 5.

If cos θ = \(\frac { 8 }{ 17 }\), then find (RBSESolutions.com) all the remaining trigonometrical ratios.

Solution.

Question 6.

If cos A = \(\frac { 5 }{ 13 }\), then find the value (RBSESolutions.com) of \(\frac { cosec\quad A }{ cos\quad A+cosec\quad A }\)

Solution.

Question 7.

If 5 tan θ = 4, then find the (RBSESolutions.com) value of \(\frac { 5sin\theta -3cos\theta }{ sin\theta +2cos\theta }\)

Solution.

Question 8.

In a ∆ABC, ∠C = 90° and if cot A = √3 and cot B = \(\frac { 1 }{ \surd 3 }\) then (RBSESolutions.com) prove that sin A cos B + cos A sin B = 1.

Solution.

Question 9.

If 16 cot A = 12, then find the (RBSESolutions.com) value of \(\frac { sinA+cosA }{ sinA-cosA }\)

Solution.

Question 10.

In figure, AD = DB and ∠B = 90° then find the (RBSESolutions.com) value of the following:

(i) sin θ

(ii) cos θ

(iii) tan θ

Solution.

Given that AD = DB and ∠B = 90°

We hope the given RBSE Solutions for Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Ex 14.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 14 Trigonometric Ratios of Acute Angles Exercise 14.1, drop a comment below and we will get back to you at the earliest.