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:

In ⊥∆ABC, ∠B = 90°
As per Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7) 2
= 576 + 49
AC2 = 625
∴ AC = 25 cm

Question 2.
In the figure, find tan P – cotR?

Solution:
In right angle ∆PQR
Using the Pythagoras theorem, we get
QR2 = PR2 – PQ2
⇒ QR2 = 132 – 122 = (13 – 12)(13 + 12) = 1 × 25 = 25
∴ QR = $$\sqrt{25}$$ = 5 cm
Now, tanP = $$\frac{Q R}{P Q}=\frac{5}{12}$$ , cotR = $$\frac{Q R}{P Q}=\frac{5}{12}$$
∴ tanP – cotR = $$\frac{5}{12}-\frac{5}{12}$$ = 0.

Question 3.
If sin A = $$\frac{3}{4}$$ calculate cos A and tan A.
Solution:

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
In the right angle triangle ABC, we have 15 cot A = 8

Question 5.
Given sec θ = $$\frac{13}{12}$$, calculate all other trigonometric ratios
Solution;

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

From equations (i) and (ii) we get:
$$\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}$$
⇒ ∆CDA ~ ∆EFB [By SSS similarity]
⇒ ∠A = ∠B Hence Proved

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:

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:

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
Solution:
In right ∆ABC, ∠B = 90°
For ∠A, we have
Base = AB, Perpendicular = BC,
Hypotenuse = AC

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

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

Question 11.
(i) The value of tan A is always less than 1.
(ii) secA = $$\frac{12}{5}$$ for some value of angle A.
(iii) cosA is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sinθ = $$\frac{4}{3}$$ for some angle θ.
Solution:
(i) False
∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.
(ii) True
∵ cos A is always less than 1.
∴ $$\frac{1}{\cos A}$$ i.e., sec A will always be greater than 1.
(iii) False
∵ ‘cosine A’ is abbreviated as ‘cosA’
(iv) False
∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.
(v) False
∵ $$\frac{4}{3}$$ is greater than 1 and sinθ cannot be greater than 1

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