**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

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

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

= 576 + 49

AC^{2} = 625

∴ AC = 25 cm

Question 2.

In the figure, find tan P – cotR?

Solution:

In right angle ∆PQR

Using the Pythagoras theorem, we get

QR^{2} = PR^{2} – PQ^{2}

⇒ QR^{2} = 13^{2} – 12^{2} = (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) cot^{2}θ

Solution:

Question 8.

If 3 cot A = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos^{2} A – sin^{2} 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

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

Question 11.

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

(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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