Students can Download Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

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

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

Answer:

In ∆^{le} ABC ∠B = 90°

AC^{2} = AB^{2} + BC^{2} [pythagoras theorem]

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

= 576 + 49 = 625

AC = \(\sqrt{625}\)

AC = 25 cm.

Question 2.

In Fig.11.3 find tan P – cot R.

Answer:

In ∆^{le} PQR ∠Q = 90°

(PR)^{2} = (PQ)^{2} + (QR)^{2} [pythagoras theorem]

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

169= 144 +(QR)^{2}

(QR)^{2} = 169 – 144

(QR)^{2} = 25

QR = \(\sqrt{25}\) = 5cm

Question 3.

If sin A = \(\frac{3}{4}\) calculate cos A and tan A

Answer:

In ∆^{le} ABC ∠B = 90°

AC^{2} = AB^{2} + BC^{2} [pythagoras theorem]

(4)^{2} = AB^{2} + (3)^{2}

16 = AB^{2} + 9

AB^{2} = 16 – 9 = 7

Question 4.

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

Answer:

15 cot A = 8

In ∆^{le} ABC ∠B = 90°

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

AC^{2}= (8)^{2} + (15)^{2}

AC^{2} = 64 + 225 – 289

Question 5.

Given sec 0 = \(\frac{13}{12}\) calculate all other trigonometric ratios.

Answer:

sec θ = \(\frac{\text { hyp }}{\text { adj }}=\frac{13}{12}\)

In ∆ ABC ∠B = 90°

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

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

AB^{2} = 169 – 144 = 25

AB = \(\sqrt{25}\)

AB = 5

Trigonometric ratios

Question 6.

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

Answer:

\(\frac{A C}{B C}\) = 1

∴ AC = BC.

In a triangle two sides are equal then opposite angles are also equal

∴ ∠A = ∠B.

Question 7.

If cot θ = \(\frac{7}{8}\) evaluate:

Answer:

In ∆^{le} ABC ∠B = 90°

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

AC^{2} = (8)^{2} + (7)^{2}

AC^{2} = 64 + 49 = 113

AC = \(\sqrt{113}\)

Question 8.

If 3 cotA = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos^{2} A – sin^{2} A or not.

Answer:

In ∆^{le} ABC ∠B = 90°

AC^{2} = AB^{2} + BC^{2} [pythagoras theorem]

AC^{2} = (4k)^{2} + (3k)^{2}

AC^{2} = 16k^{2} + 9k^{2}

AC^{2} = 25k^{2}

AC = \(\sqrt{25 k^{2}}\) = 5k.

AC = 5k.

∴ LHS = RHS

\(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos^{2}A – sin^{2}A

Question 9.

In triangle ABC, right-angled at B, if tan A = \(\frac{1}{\sqrt{3}}\), find the value of:

i) sin A cos C + cos A sin C

ii) cos A cos C – sin A sin C

Answer:

In ∆^{le} ABC ∠B = 90°

AC^{2} = AB^{2} + BC^{2} [pythagoras theorem]

AC^{2} = (\(\sqrt{3}\)k)^{2} + (1k)^{2}

AC^{2} = 3k^{2} + k^{2} = 4k^{2}

AC^{2} = \(\sqrt{4}\)k^{2} = 2k

i) sin A cos C + cos A sin C

ii) cos A cos C – sin A sin C

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. Determine the values of sin P, cos P and tan P.

Answer:

PR + QR = 25 cm and PQ = 5 cm

Let PR = x cm and QR = (25 – x)

In ∆^{le} PQR ∠Q = 90°

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

(x)^{2} = (5)^{2} + (25 – x)^{2}

x^{2} = 25 + 625 – 50x + x^{2}

x^{2} – x^{2} – 50x + 650 = 0

50x = 650

x = \(\frac{650}{50}\)

x = 13 cm

PR = x = 13cm and QR = 25 – x = 25 – 13 = 12 cm, QR = 12 cm