KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 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
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 1
In ⊥∆ABC, ∠B = 90°
As per Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7) 2
= 576 + 49
AC2 = 625
∴ AC = 25 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 2

Question 2.
In the given figure, find tan P – cot R.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 3
In ⊥∆PQR, ∠Q = 90°
∴ PQ2 + QR2 = PR2
(12)2 + QR2 = (13)2
144 + QR2 = 169
QR2 = 169 – 144
QR2 = 25
∴ QR = 5 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 4

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 5
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 6

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 7

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios
Solution;
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 8

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 9
CD is common.
S.S.S. Postulate :
∴ ∆ADC ~ ∆ADB
∴ ∠A = ∠B
∵ “Angles of similar triangles are equiangular.”

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate :
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot2θ
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 10
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 11

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos2 A – sin2 A or not
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 12
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 13

Question 9.
In ∆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
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 14
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 15

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
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 16
PR2 = PQ2 + QR2
QR2 = PR2 – PQ2
= (25 – QR)2 – (5)2
QR2 = 625 – 50QR + QR2 – 25
50QR = 600
∴ QR = 12 cm.
∴ PR = 25 – QR = 25 – 12 = 13 cm.
∴ QR = 12 cm
∴ PR = 25 – QR = 25 – 12 = 13 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 17

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° = \(\sqrt{3}\) > 1

(ii) If sec A = \(\frac{12}{5}\) 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 = \(\frac{\text { Adjacent side }}{\text { Opposite side }}\)

(v) sin θ = \(\frac{4}{3}\) for some angle θ
Answer:
False. because sin θ ≯ 1

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1, drop a comment below and we will get back to you at the earliest.

Leave a Comment

error: Content is protected !!