RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

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
AC² = AB² + BC²
AC² = (24)² + (7)²
AC² = 576 + 49
AC² = – 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
PR² = PQ² + QR²
(13)² = (12)² + QR²
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
AC² = AB² + BC²
or (4K)² = (AB)² + (3K)²
or 16K² = AB² + 9K²
or 16K² – 9K² = AB²
or 7K² = AB²
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
AC² = (AB)² + (BC)²
(AC)² = (8K)² + (15K)²
(AC)² = 64K² + 225K²
(AC)² = 289K²
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 AC² = (AB)² + (BC)²
or (13k)² = (12k)² + (BC)²
or 169k² = 144k² + (BC)²
or 169k² = 144*2 = (BC)²
or (BC)² = 25k²
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) cot²θ
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
AB² = (BC)² + (AC)²
or (AB)² = (7k)² + (8k)²
or (AB)² = 49k² + 64k²
or (AB)² = 113k²
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 \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos² A – sin² 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 = 16k² + 9k²
(AC)² = 25k²
AC = ±\(\sqrt{25 k^{2}}\)
AC = ± 5k
But AC = 5k
[AC ≠ – 5k, since side cannot be negative]



∴ cos²A – sin²A = \(\frac{7}{25}\)
From (i) and (ii),
L.H.S. = R.H.S.
∴ \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos²A – sin²A

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 AC² = 3k² + k²
or AC² = 4k²
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.