SEBA Class 10 Maths Chapter 8 Introduction to Trigonometry

SEBA Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 8 Introduction to Trigonometry Notes and select need one. SEBA Class 10 Maths Chapter 8 Introduction to Trigonometry Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 8 Introduction to Trigonometry

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Introduction to Trigonometry

Chapter: 8

Exercise 8.1

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

(i) sin A, cosA

(ii) sin C, cosC

Ans: Applying Pythagoras theorem for ∆ABC, we obtain

AC² = AB² + BC²

= (24 cm)² + (7 cm)²

= (576 + 49) cm²

= 625 cm²

∴ AC = cm = 25 cm

2. In fig. 8.13 find tan P − cot R. 

Ans: Applying Pythagoras theorem for ∆PQR, we obtain

PR² = PQ² + QR²

(13 cm)² = (12 cm)² + QR²

169 cm² = 144 cm² + QR²

25 cm² = QR²

QR = 5 cm

3. If sin A = 3/4, calculate cos A and tan A.

Ans: 

4. Given 15 cotA = 8, find sinA and sec A.

Ans: Consider a right-angled triangle, right-angled at B.  

5. Given secθ = 13/12, calculate all other trigonometric ratios.

Ans:  

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

(AC)² = (AB)² + (BC)²

(13k)² = (12k)² + (BC)²

169k² = 144k² + BC²

25k² = BC²

BC = 5k

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

Ans: Let us consider a triangle ABC in which CD ⊥ AB.

It is given that cosA = cosB

AB/AC = BD/BC …(1)

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.

7.  If cot θ = 7/8  evaluate 

Ans: Let us consider a right triangle ABC, right-angled at point B.

If BC is 7k, then AB will be 8k, where k is a positive integer. Applying Pythagoras theorem in ∆ABC, we obtain  

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k² 

(ii) cot² θ

Ans:

Ans: If 3 cot A = 4, 

It is given that 3 cot = 3/4 A = 4 Or, cot A = Consider a right triangle ABC, right-angled at point B. 

If AB is 4k, then BC will be 3k, where k is a positive integer.

In ∆ABC,

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

AC = 5k 

9. In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:

(i) sinA cosC + cosA sinC

(ii) cosA cosC – sinA sinC

10. In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.

Ans: Given that, PR + QR = 25 cm

 PQ = 5 cm

Let PR be x. 

Therefore, QR = (25 − x)cm

Applying Pythagoras theorem in ∆PQR, we obtain

PR² = PQ² + QR²

x² = (5)² + (25 − x)²

x² = 25 + 625 + x² − 50x

50x = 650

x = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

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

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

Ans: (i) Consider a ∆ABC, right-angled at B.

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii) Sec A = 12/5 for some value of angle A.

Ans:

Let AC be 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ∆ABC, we obtain

AC² = AB² + BC²

(12k)² = (5k)² + BC²

144k² = 25k² + BC²

BC² = 119k^2.         

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec

A is possible.

Hence, the given statement is true. 

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

Ans: The abbreviation used for cosecant of angle A is cosec A. And cos A is the

abbreviation used for the cosine of angle A.

Hence, the given statement is false. 

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

Ans: cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false.

(v) sin θ = 4/3, for some angle θ

Ans: sin θ = 4/3 We know that in a right-angled triangle,

In a right-angled triangle, the hypotenuse is always greater than the remaining two sides.

Therefore, such a value of sin θ is not possible.

Hence, the given statement is false.

12. Express the ratios sin A tan A cosecА, cotA in terms of cos A.

Ans: 

13. In a triangle ABC, ∠B = 90°, sin A = 3/5 then the value of cos A is:

(a) 5/3

(b) 3/5

(c) 4/5

(d) 5/4

Ans: (c) 4/5

14. In ΔPQR, right angle at Q, ∠P = 45°, PQ = 7 cm, then length of QR is:

(a) 7√2 cm

(b) 14 cm

(c) √7 cm

(d) 7 cm

Ans: (d) 7 cm