Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.

In △ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Determine :

(i) sin A, cos A

(ii) sin C, cos C.

Solution:

(i) To determine—sin A, cos A

Here AB = 24 cm; BC = 7 cm

By Pythagoras Therem

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

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

AC^{2} = 576 + 49

AC^{2} = – 625

AC = \(\sqrt {625}\)

AC = 25 cm.

∴ sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

∴ sin A = \(\frac{7 \mathrm{~cm}}{25 \mathrm{~cm}}\) = \(\frac{7}{25}\)

Question 2.

In Fig. find tan P – cot R.

Solution:

Hypotenuse PR = 13 cm

By Pythagoras Theorem

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

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

or 169 = 144 + (QR)^{2}

or 169 – 144 = (QR)^{2}

or 25 = (QR)^{2}

or QR = ±\(\sqrt {25}\)

or QR = 5,-5

But QR= 5 cm

[QR ≠ -5 since the side cannot be negative]

Question 3.

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

Solution:

Let ABC be any right angled triangle in which ∠B is a right angle.

sin A = \(\frac{3}{4}\)

Let BC = 3K

AC = 4K

By Pythagoras Theorem

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

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

or 16K^{2} = AB^{2} + 9K^{2}

or 16K^{2} – 9K^{2} = AB^{2}

or 7K^{2} = AB^{2}

AB = ±\(\sqrt{7 \mathrm{~K}^{2}}\)

AB= ±\(\sqrt {7}\)K

[AB ≠ –\(\sqrt {7}\) K since side cannot be negative]

⇒ AB = \(\sqrt {7}\)K

Question 4.

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

Solution:

Let ABC be any right angled triangle in which A is an acute angle and B is a right angle.

15 cot A = 8

By Pythagoras Theorem

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

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

(AC)^{2} = 64K^{2} + 225K^{2}

(AC)^{2} = 289K^{2}

AC = ±\(\sqrt{289 \mathrm{~K}^{2}}\)

AC = ± 17K

⇒ AC = 17K

[AC = – 17K, since side cannot be negative]

Question 5.

Given sec 9 = \(\frac{13}{12}\) calculate all other trigonometric ratios.

Solution:

Let ABC be any right angled triangle in which B is a right angle.

Again let ∠BAC = θ

According to the question

By Pythagoras Theorem

or AC^{2} = (AB)^{2} + (BC)^{2}

or (13k)^{2} = (12k)^{2} + (BC)^{2}

or 169k^{2} = 144k^{2} + (BC)^{2}

or 169k^{2} = 144*2 = (BC)^{2}

or (BC)^{2} = 25k^{2}

or BC = ±\(\sqrt{25 \mathrm{~k}^{2}}\)

or BC = ± 5k

or BC = 5k.

[BC ≠ – 5k since side cannot be negative]

Question 6.

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

Solution:

Let ABC be any triangle where ∠A and ∠B arc acute angles. We arc to determine cos A and cos B. Now draw CM ⊥ AB.

∴ ∠AMC = ∠BMC = 90°

In right △AMC

\(\frac{\mathrm{AM}}{\mathrm{AC}}\) = cos A …..(i)

Again in right △BMC.

\(\frac{\mathrm{BM}}{\mathrm{BC}}\) = cos B …….(ii)

But cos A= cos B (Given) …..(iii)

From (i). (ii) and (iii),

⇒ \(\frac{\mathrm{AM}}{\mathrm{AC}}\) = \(\frac{\mathrm{BM}}{\mathrm{BC}}\)

\(\frac{\mathrm{AM}}{\mathrm{AC}}\) = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{\mathrm{CM}}{\mathrm{CM}}\)

∴ △AMC ~ △BMC

[By SSS similarity criteria]

⇒ ∠A = ∠B

[∵ Corresponding angles of similar triangles are equal.]

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:

(i) ∠ABC = 0 In right angled triangle ACB, there is right angle at C.

According to the question

cot θ = \(\frac{7}{8}\)

But cot θ = \(\frac{BC}{AC}\)

⇒ \(\frac{BC}{AC}\) = \(\frac{7}{8}\)

Lct BC = 7k, AC = 8k

By Pythagoras Theorem

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

or (AB)^{2} = (7k)^{2} + (8k)^{2}

or (AB)^{2} = 49k^{2} + 64k^{2}

or (AB)^{2} = 113k^{2}

or AB = ±\(\sqrt{113 k^{2}}\)

or AB = \(\sqrt {113}\) k

[AB ≠ –\(\sqrt {113}\) k since side cannot be negative]

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:

Let ABC be a right angled triangle in which there is a right angle at B.

According to the question

3 cot A = 4

∴ cot A = \(\frac{4}{3}\)

But cot A = \(\frac{AB}{BC}\)

⇒ \(\frac{AB}{BC}\) = \(\frac{4}{3}\)

Let AB = 4k, BC = 3k

By Pythagoras Theorem

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

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

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

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

AC = ±\(\sqrt{25 k^{2}}\)

AC = ± 5k

But AC = 5k

[AC ≠ – 5k, since side cannot be negative]

∴ cos^{2}A – sin^{2}A = \(\frac{7}{25}\)

From (i) and (ii),

L.H.S. = R.H.S.

∴ \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = 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.

Solution:

(i) According to the question

△ABC is a right angled triangle where angle B is a right angle.

tan A = \(\frac{1}{\sqrt{3}}\) ……(i)

But tan A = \(\frac{BC}{AB}\) ……….(ii)

From (i) and (ii),

\(\frac{BC}{AB}\) = \(\frac{1}{\sqrt{3}}\)

Let BC = k, AB = \(\sqrt {3}\) k

By Pythagoras Theorem,

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

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

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

or AC^{2} = 4k^{2}

or AC = ±\(\sqrt{4 k^{2}}\)

AC = ± 2k

AC = 2k

[AC ≠ – 2k V side cannot be negative]

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:

According to the question

In △PQR, there is a right angle at Question

PR + QR = 25 cm

PQ = 5 cm

In right angled triangle PQR,

By Pythagoras Theorem,

(PR)^{2} = (PQ)^{2} + (RQ)^{2}

or (PR)^{2} = (5)^{2} + (RQ)^{2}

[∴ PR + QR = 25

∵ QR = 25 – PR]

or (PR)^{2} = 25 + [25 – PR]^{2}

or (PR)^{2} = 25 + (25)^{2} + (PR)^{2} – 2 × 25 × PR

or (PR)^{2} = 25 + 625 + (PR)^{2} – 50PR

or (PR)^{2} – (PR)^{2} + 50PR = 650

or 50PR = 650

or PR = \(\frac{650}{50}\)

PR = 13

QR = 25 – PR

⇒ QR= 25 – 13

QR = 12 cm

sin P = \(\frac{QR}{PR}\) = \(\frac{12}{13}\)

cos P = \(\frac{PQ}{PR}\) = \(\frac{5}{13}\)

tan P = \(\frac{QR}{PQ}\) = \(\frac{12}{5}\)

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) sec A = \(\frac{12}{5}\) for some value of angle A.

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

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

(v) sin θ = \(\frac{4}{3}\) for some angle θ.

Solution:

(i) False.

Since nothing can be said about the lengths of the sides of right angled triangle. Therefore tan A mav assume any value.

(ii) sec A = \(\frac{,hypotenuse}{base}\) = \(\frac{12}{5}\) = 2.40 >1

∵ Ratio of hypotenuse and base is 12 : 5. It is not necessary that it so happens always, but for some value it is true.

Hence the statement is true.

(iii) ∵ cos A is the short form of cosine A of angle A whereas cosecant A means cosec A. Hence the given statement is false.

(iv) False.

Since cot A means cotangent of angle A not the product of cot and A.

(v) False.

sin θ = \(\frac{4}{3}\) = 1.333 > 1

Since the value of sin θ is always 1 and less than 1.

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