In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 pdf, free KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 10 Maths Solutions** **Chapter 11 Introduction to Trigonometry Ex 11.1**

**Trigonometry Exercise 11.1 Solutions Question 1.**

In ∆ABC, right-angled at B, AB = 24 cm., BC = 7 cm. Determine:

i) sin A, cos A

ii) sin C, cos C

Solution:

In ⊥∆ABC, ∠B = 90°

As per Pythagoras theorem

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

= (24)^{2} + (7) ^{2}

= 576 + 49

AC^{2} = 625

∴ AC = 25 cm

**Introduction To Trigonometry Class 10 Exercise 11.1 Question 2.**

In the given figure, find tan P – cot R.

Solution:

In ⊥∆PQR, ∠Q = 90°

∴ PQ^{2} + QR^{2} = PR^{2}

(12)^{2} + QR^{2} = (13)^{2}

144 + QR^{2} = 169

QR^{2} = 169 – 144

QR^{2} = 25

∴ QR = 5 cm.

**Introduction To Trigonometry Exercise 11.1 Question 3.**

If sin A = 34 calculate cos A and tan A.

Solution:

**Exercise 11.1 Class 10 Trigonometry Question 4.**

Given 15 cot A = 8, find sin A and sec A.

Solution:

**KSEEB Solutions For Class 10 Maths Trigonometry Question 5.**

Given sec θ = 1312, calculate all other trigonometric ratios

Solution;

**10th Maths Introduction To Trigonometry Exercise 11.1 Question 6.**

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:

In ⊥∆ABC, ∠A and ∠B are acute angles. CD ⊥ AB is drawn.

CD is common.

S.S.S. Postulate :

∴ ∆ADC ~ ∆ADB

∴ ∠A = ∠B

∵ “Angles of similar triangles are equiangular.”

**10th Class Maths Trigonometry Exercise 11.1 Question 7.**

If cot θ = 78, evaluate :

(i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)

(ii) cot^{2}θ

Solution:

**Trigonometry Class 10 Exercise 11.1 Question 8.**

If 3 cot A = 4, check whether 1−tan2A1+tan2A = cos^{2} A – sin^{2} A or not

Solution:

**Introduction To Trigonometry Class 10 KSEEB Question 9.**

In ∆ABC, right-angled at B, if tan A =13√ find the value of:

i) sin A cos C + cos A sin C

ii) cos A cos C – sin A sin C

Solution:

**Trigonometry Ex 11.1 Question 10.**

In ∆PQR, right-angled at Q, PR + QR = 25 cm. and PQ = 5 cm. Determine the values of sin P, cos P and tan P

Solution:

PQ = 5 cm

PR + QR = 25 cm

∴ PR = 25 – QR

PR^{2} = PQ^{2} + QR^{2}

QR^{2} = PR^{2} – PQ^{2}

= (25 – QR)^{2} – (5)^{2}

QR^{2} = 625 – 50QR + QR^{2} – 25

50QR = 600

∴ QR = 12 cm.

∴ PR = 25 – QR = 25 – 12 = 13 cm.

∴ QR = 12 cm

∴ PR = 25 – QR = 25 – 12 = 13 cm

**10th Class Trigonometry Exercise 11.1 Question 11.**

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

Answer:

False. 60° = 3–√ > 1

(ii) If sec A = 125 for some value of angle A.

Answer:

True. because sec A > 1.

(iii) cos A is the abbreviation used for the cosecant of angle A.

Answer:

False. Because cos A is simplified as cos.

(iv) cot A is the product of cot and A.

Answer:

False. cot is ∠A or meaningless. Here,

cot A = Adjacent side Opposite side

(v) sin θ = 43 for some angle θ

Answer:

False. because sin θ ≯ 1

**All Chapter KSEEB Solutions For Class 10 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

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