In this chapter, we provide RD Sharma Class 10 Ex 5.1 Solutions Chapter 5 Trigonometric Ratios for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 5.1 Solutions Chapter 5 Trigonometric Ratios pdf, Now you will get step by step solution to each question.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 5 |
| Chapter Name | Trigonometric Ratios |
| Exercise | 5.1 |
| Category | RD Sharma Solutions |
RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Ex 5.1 Download PDF
Chapter 5: Trigonometric Ratios Exercise – 5.1
Question: 1
Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given.
(i) sin A = 2/3
(ii) cos A = 4/5
(iii) tan θ = 11/1
(iv) sin θ = 11/15
(v) tan α = 5/12
(vi) sin θ = √3/2
(vii) cos θ = 7/25
(viii) tan θ = 8/15
(ix) cot θ = 12/5
(x) sec θ = 13/5
(xi) cosec θ = √10
(xii) cos θ =12/15
Solution:
(i) sin A = 2/3
Given: sin A = 2/3 … (1)
By definition,
By Comparing (1) and (2)
We get,
Perpendicular side = 2 and Hypotenuse = 3
Therefore, by Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Therefore,
32 = AB2 + 22
AB2 = 32 – 22
AB2 = 9 – 4
AB2 = 5
AB = √5
Hence, Base = √5
Therefore,
Therefore,
Therefore,
cot A = √5/2
(ii) cos A = 4/5
Given: cos A = 4/5 … (1)
By Definition,
By comparing (1) and (2)
We get,
Base = 4 and
Hypotenuse = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side (BC)
52 = 42+ BC2
BC2 = 52 – 42
BC2 = 25 – 16
BC2 = 9
BC = 3
Hence, Perpendicular side = 3
Now,
Therefore,
sin A = 3/5
Now, cosec A = 1/(sin A)
Therefore,
cosec A = 1/(sin A)
Therefore,
Therefore,
Therefore,
tan A = 3/4
Now, cot A = 1/(tan A)
Therefore,
By definition,
By Comparing (1) and (2)
We get,
Base = 1 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse (AC)
AC2 = 12 + 112
AC2 = 1 + 121
AC2 = 122
AC = √122
Therefore,
Sin θ = 11√122
Therefore,
Therefore,
Therefore,
By definition,
By Comparing (1) and (2)
We get,
Perpendicular Side = 11 and
Hypotenuse = 15
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
152 = AB2 +112
AB2 = 152 – 112
AB2= 225 – 121
AB2 = 104
Therefore,
Therefore,
Therefore,
Therefore,
Therefore,
(v) tan α = 5/12
Given: tan α = 5/12 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC)
AC2 = 122 + 52
AC2 = 144 + 25
AC2 = 169
AC = 13
Hence Hypotenuse = 13
Therefore,
sin α = 5/13
Therefore,
Therefore,
By definition,
By comparing (1) and (2)
We get,
Perpendicular side = √3
Hypotenuse = 2
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
AB2 = 4 – 3
AB2 = 1
AB = 1
Hence Base = 1
Therefore,
cos θ = 1/2
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
Therefore,
Therefore,
(vii) cos θ = 7/25
Given: cos θ = 7/25 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 7 and
Hypotenuse = 25
Therefore
By Pythagoras theorem,
AC 2 = AB2 + BC2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
252 = 72 + BC2
BC2 = 252 – 72
BC2 = 625 – 49
BC = 576
BC = √576
BC = 24
Hence, Perpendicular side = 24
Therefore,
Sin θ = 24/25
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
Therefore,
Therefore,
By definition,
By comparing (1) and (2)
We get,
Base = 15 and Perpendicular side = 8
Therefore,
By Pythagoras theorem,
AC2= 152 + 82
AC2= 225 + 64
AC2 = 289
AC = √289
AC = 17
Hence, Hypotenuse = 17
Therefore,
sin θ = 8/17
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
cos θ =15/17
Now, sec θ = 1/cos θ
Therefore,
Therefore,
(ix) cot θ = 12/5
Given: cot θ = 12/5 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and perpendicular side (BC) and get the hypotenuse (AC)
AC2 = 122 + 52
AC2 = 144 + 25
AC2 = 169
AC = √169
AC = 13
Hence, Hypotenuse = 13
Therefore,
sin θ = 5/13
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
cos θ = 12/13
Now, sec θ = 1/cos θ
Therefore,
Therefore,
(x) sec θ = 13/5
Given: sec θ = 13/5… (1)
By definition,
By comparing (1) and (2)
We get,
Base = 5
Hypotenuse = 13
Therefore,
By Pythagoras theorem,
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
132 = 52 + BC2
BC2 = 132 – 52
BC2 = 169 – 25
BC2 = 144
BC = √144
BC = 12
Hence, Perpendicular side = 12
Therefore,
sin θ = 12/13
Now, cosec θ = 1/sin θ
Therefore,
Therefore,
Therefore,
tan θ = 12/5
Now, cot θ = 1/tan θ
Therefore,
(xi) cosec θ = √10
Given: cosec θ = √10/1 … (1)
By definition
By comparing (1) and (2)
We get,
Perpendicular side = 1 and
Hypotenuse = √10
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
AB2 = 10 – 1
AB = √9
AB = 3
Hence, Base side = 3
Therefore,
Therefore,
Therefore,
cot θ = 3
(xii) cos θ = 12/15
Given: cos θ = 12/15 … (1)
By definition,
By comparing (1) and (2)
We get,
Base = 12 and Hypotenuse = 15
Therefore,
By Pythagoras theorem,
AC2 = AB2 + BC2
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
152 = 122 + BC2
BC2 = 152 – 122
BC2 = 225 – 144
BC 2= 81
BC = √81
BC = 9
Hence, Perpendicular side = 9
Therefore,
Therefore,
Therefore,
Therefore,
Therefore,
Question: 2
In a ΔABC, right angled at B, AB – 24 cm, BC = 7cm, Determine
(i) sin A, cos A
(ii) sin C, cos C
Solution:
(i) The given triangle is below:
Given: In ΔABC, AB = 24 cm
BC = 7cm
∠ABC = 90°
To find: sin A, cos A
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC2 = 625
AC = √625
AC = 25
Hypotenuse = 25
By definition,
By definition,
Answer:
(ii) The given triangle is below:
Given: In Δ ABC, AB = 24 cm
BC = 7cm
∠ABC = 90°
To find: sin C, cos C
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC2 = 625
AC = √625
AC = 25
Hypotenuse = 25
By definition,
By definition,
By definition,
Question: 3
In the below figure, find tan P and cot R. Is tan P = cot R? To find, tan P, cot R
Solution:
In the given right angled ΔPQR, length of side OR is unknown
Therefore, by applying Pythagoras theorem in ΔPQR
We get,
PR2 = PQ2 + QR2
Substituting the length of given side PR and PQ in the above equation
132 = 122 + QR2
QR2 = 132 – 122
QR2 = 169 – 144
QR2= 25
QR =√25
By definition, we know that,
Also, by definition, we know that
Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.
Therefore, L.H.S of both equations is also equal
tan P = cot R
Answer:
Question: 4
If sin A = 9/41, Compute cos A and tan A.
Solution:
Given: sin A = 9/41 … (1)
To find: cos A, tan A
By definition,
By comparing (1) and (2)
We get,
Perpendicular side = 9 and Hypotenuse = 41
Now using the perpendicular side and hypotenuse we can construct ΔABC as shown figure.
Length of side AB is unknown is right angled ΔABC,
To find the length of side AB, we use Pythagoras theorem,
Therefore, by applying Pythagoras theorem in ΔABC,
We get,
AC2 = AB2 + BC2
412 = AB2 + 92
AB2 = 412 – 92
AB2 = 168 – 81
AB= 1600
AB = √1600
AB = 40
Hence, length of side AB = 40
Now
By definition,
Now,
By definition,
Question: 5
Given 15 cot A = 8, find sin A and sec A.
Solution:
Given: 15 cot A = 8
To find: sin A, sec A
Since 15 cot A =8
By taking 15 on R.H.S
We get,
Since 15 cot A = 8
By taking 15 on R.H.S
We get,
cot A = 8/15
By definition,
cot A = 1/(tan A)
Hence,
Comparing equation (1) and (2)
We get,
Base side adjacent to ∠A = 8
Perpendicular side opposite to ∠A = 15
ΔABC can be drawn below using above information
Hypotenuse side is unknown.
Therefore, we find side AC of ΔABC by Pythagoras theorem.
So, by applying Pythagoras theorem to ΔABC
We get,
AC2 = AB2 +BC2
Substituting values of sides from the above figure
AC2 = 82 + 152
AC2 = 64 + 225
AC2 = 289
AC = √289
AC = 17
Therefore, hypotenuse =17
Now by definition,
Substituting values of sides from the above figure
Sin A = 15/17
By definition,
sec A = 1/cos A
Hence,
Substituting values of sides from the above figure
Sec A = 17/8
Answer:
sin A =15/17, sec A = 17/8
Question: 6
In ΔPQR, right angled at Q, PQ = 4cm and RQ = 3 cm .Find the value of sin P, sin R, sec P and sec R.
Solution:
Given: ΔPQR is right angled at vertex Q.
PQ = 4cm
RQ = 3cm
To find,
sin P, sin R, sec P, sec R
Given ΔPQR is as shown figure.
Hypotenuse side PR is unknown.
Therefore, we find side PR of ΔPQR by Pythagoras theorem
By applying Pythagoras theorem to ΔPQR
We get,
PR2 = PQ2 + RQ2
Substituting values of sides from the above figure
PR2 = 42 +32
PR2 = 16 + 9
PR2 = 25
PR = √25
PR = 5
Hence, Hypotenuse =5
Now by definition,
sin P = RQ/PR
Substituting values of sides from the above figure
sin P = 3/5
Now by definition,
sin R = PQ/PR
Substituting the values of sides from above figure
sin R = 4/5
By definition,
sec P = 1/cos P
Substituting values of sides from the above figure
sec P = PR/PQ
sec P = 5/4
By definition,
sec R = 1/cos R
Substituting values of sides from the above figure
sec R = PR/RQ
sec R = 5/3
Answer:
sin P = 3/5,
sin R = 4/5,
sec P = 5/4,
sec R = 5/3
Question: 7
If cot θ = 7/8, evaluate
(ii) cot2 θ cot2θ
Solution:
Given: cot θ = 7/8
We know the following formula
(a + b)(a – b) = a2 – b2
By applying the above formula in the numerator of equation (1)
We get,
(1 + sin θ) × (1 – sin θ) = 1 – sin2θ …. (2) (Where, a = 1 and b = sin θ)
Similarly,
By applying formula (a + b) (a – b) = a2 – b2 in the denominator of equation (1).
We get,
(1 + cos θ)(1 – cos θ) = 1 – cos2 θ (1 + cos θ)(1 – cos θ) = 1 – cos2θ … (Where a = 1 and b = cos θ cos θ
Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).
Therefore,
Since,
cos2θ + sin2 θ = 1 cos2 θ + sin2 θ = 1
Therefore,
cos2 θ = 1 – sin2Θcos2 θ =1 – sin2 θ
Also, sin2 θ = 1 – cos2 θ sin2 θ = 1 – cos2 θ
Putting the value of 1 – sin2 θ and 1 – cos2 θ in equation (4)
We get,
We know that,
Since, it is given that cot θ = 7/8
Therefore,
(ii) Given: cot θ = 7/8
To evaluate: cot2 θ
cot θ =7/8
Squaring on both sides,
We get,
Answer:
49/64
Question: 8
If 3 cot A = 43 cot A = 4, check whether
Solution:
Given: 3 cot A = 4
To check whether
cos2A –sin2A or not.
3 cot A = 4
Dividing by 3 on both sides,
We get,
cot A = 4/3 … (1)
By definition,
cot A = 1/tan A
Therefore,
Comparing (1) and (2)
We get,
Base side adjacent to ∠A = 4
Perpendicular side opposite to ∠A = 3
Hence ΔABC is as shown in figure.
In ΔABC , Hypotenuse is unknown
Hence, it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem in ΔABC
We get
AC2 = AB2 + BC2
Substituting the values of sides from the above figure
AC2 = 42 + 32
AC2 = 16 + 9
AC2 = 25
AC = √25
AC = 5
Hence, hypotenuse = 5
To check whethercos2A – sin2A or not.
We get thee values of tan A, cos A, sin A
By definition,
tan A =1/(cot A)
Substituting the value of cot A from equation (1)
We get,
tan A = 1/4
tan A = 3/4 …. (3)
Now by definition,
Substituting the values of sides from the above figure
Now we first take L.H.S of equation
Substituting value of tan A from equation (3)
We get,
Taking L.C.M on both numerator and denominator
We get,
Now we take R.H.S of equation whether
R.H.S = cos2A – sin2A
Substituting value of sin A and cos A from equation (4) and (5)
We get,
Comparing (6) and (7)
We get.
Answer:
Question: 9
If tan θ = a/b, find the value of
Solution:
Given: tan θ = a/b … (1)
Now, we know that
Therefore equation (1) become as follows
Now, by applying invertendo
We get,
Now by applying Componendo – dividendo
We get,
Therefore,
Question: 10
If 3 tan θ = 4, find the value of
Solution:
Given: If 3 tan θ = 4/3 tan θ = 4
Therefore,
tan θ = 4/3 … (1)
Now, we know that
Therefore equation (1) becomes
Now, by applying Invertendo to equation (2)
We get,
Now, multiplying by 4 on both sides
We get
Therefore
Now, multiplying by 2 on both sides of equation (3)
We get,
Now by applying componendo in above equation
We get,
Therefore,
Therefore, on L.H.S sin θ sin θ cancels and we get,
Therefore,
4 cos θ – sin θ = 44 cos θ – sin θ = 4
Question: 11
If 3 cot θ = 2, find the value of
Solution:
Given: 3 cot θ = 2
Therefore,
Now, we know that
Therefore equation (1) becomes
Now, by applying invertendo to equation (2)
Now, multiplying by 4/3 on both sides,
We get,
Therefore, 3 cancels out on R.H.S and
We get,
Now by applying invertendo dividendo in above equation
We get,
Now, multiplying by 2/6 on both sides of equation (3)
We get,
Therefore, 2 cancels out on R.H.S and
We get,
Now by applying componendo in above equation
We get,
Now, by dividing equation (4) by (5)
We get,
Therefore,
Therefore, on L.H.S (3 sin θ) cancels out and we get,
Now, by taking 2 in the numerator of L.H.S on the R.H.S
We get,
Therefore, 2 cancels out on R.H.S and
We get,
Hence answer,
Question: 12
If tan θ = a/b, prove that
Solution:
Given:
Now, we know that
Therefore equation (1) becomes
Now, by multiplying by a/b on both sides of equation (2)
We get,
Therefore,
Now by applying dividendo in above equation (3)
We get,
Now by applying componendo in equation (4)
We get,
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, b cos θ and b2 cancels on L.H.S and R.H.S respectively
Hence, it is proved that
Question: 13
If sec θ = 13/5, show that
Solution:
Given: sec θ = 13/5
To show that
Now, we know that
Therefore,
Therefore,
Now, we know that
Now, by comparing equation (1) and (2)
We get,
Base side adjacent to ∠θ = 5
And
Hypotenuse =13
Therefore from above figure
Base side BC = 5
Hypotenuse AC = 13
Side AB is unknown. It can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2 = AB2 + BC2
Therefore by substituting the values of known sides
We get,
132 = AB2 + 52
Therefore,
AB2= 132 – 52
AB2= 169 – 25
AB2 = 144
AB = √144
Therefore,
AB = 12 …. (3)
Now, we know that
sin θ = AB/AC
sin θ = 12/13 … (4)
Now L.H.S of the equation to be proved is as follows
Substituting the value cos θ of sin θ and from equation (1) and (4) respectively
We get,
Therefore,
L.H.S = 3
Hence proved that,
Question: 14
If cos θ = 12/13, show that sin θ(1 – tan θ) = 35/156
Solution:
To show that
Now we know that
Therefore, by comparing equation (1) and (2)
We get,
Base side adjacent to ∠θ = 12
And
Hypotenuse = 13
Therefore from above figure
Base side BC = 12
Hypotenuse AC = 13
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values of known sides
We get,
132= AB2 + 122
Therefore,
AB2= 132 – 122
AB2= 169 – 144
AB = 25
AB = √25
AB = 5 …. (3)
Now, we know that
Now from figure (a)
We get,
sin θ = AB/AC
Therefore,
sin θ = 5/12… (5)
Now L.H.S of the equation to be proved is as follows
L.H.S of the equation to be proved is as follows
L.H.S = sin θ (1 – tan θ] …. (6)
Substituting the value of sin θ and tan θ from equation (4) and (5)
We get,
Taking L.C.M inside the bracket
We get,
Therefore,
Now by opening the bracket and simplifying
We get,
From equation (6) and (7),it can be shown that
Question: 15
Solution:
Now, we know that
Therefore,
Therefore,
Comparing Equation (1) and (2)
We get.
Base side adjacent to ∠θ = 1
Perpendicular side opposite to ∠θ = √3
Therefore, triangle representing angle √3 is as shown below
Therefore, by substituting the values of known sides
We get,
Therefore,
AC2 = 3 + 1
AC2 = 4
AC = √4
Therefore,
AC = 2 … (3)
Now, we know that
Now from figure (a)
Therefore from figure (a) and equation (3),
Now we know that
Now from figure (a)
We get, BC/AC
Therefore from figure (a) and equation (3),
Now, L.H.S of the equation to be proved is as follows
Substituting the value of from equation (4) and (5)
We get,
Now by taking L.C.M in numerator as well as denominator
We get,
Therefore,
Therefore,
Question: 16
Solution:
To show that
Now, we know that
Therefore,
Comparing equation (1) and (2)
We get.
Perpendicular side opposite to ∠θ = 1
Base side adjacent to ∠θ = √7
Therefore, Triangle representing ∠ θ is shown figure.
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2 = AB2 + BC2
Therefore by substituting the values of known sides
We get,
Therefore,
AC 2 = 1 +7
AC2 = 8
AC = √8
Therefore,
Now we know that
Now, we know that
Therefore, from equation (4)
We get,
Now, we know that
Now from figure (a)
We get,
cos θ = BC/AC
Therefore from figure (a) and equation (3)
Now we know that
Therefore, from equation (6)
We get,
Now, L.H.S of the equation to be proved is as follows
Substituting the value of cosec θ and sec θ from equation (6) and (7)
We get,
Therefore,
Therefore,
L.H.S = 48/64
L.H.S = 3/4 = R.H.S
Therefore,
Hence proved that
Question: 17
If sec θ = 5/4, find the value of
Solution:
To find the value of
Now we know that
Therefore,
Therefore from equation (1)
Also, we know that cos2θ + sin2θ = 1
Therefore,
Substituting the value of cos θ cos θ from equation (2)
We get,
Therefore,
Also, we know that
sec2θ =1 + tan2 θ sec2 θ = 1 + tan2 θ
Therefore,
Therefore,
tan θ = 3/4 … (4)
Also, cot θ = 1/tan θ
Therefore from equation (4)
We get,
cot θ = 4/3 … (5)
Substituting the value of cos θ, cot θ and tan θ from the equation (2), (3), (4) and (5) respectively in the expression below
We get,
= 12/7
Therefore,
Question: 18
If sin θ = 12/13, find the value of
Solution:
To find the value of
Now, we know the following trigonometric identity
cosec2 θ = 1 + tan2θ
Therefore, by substituting the value of tan θ from equation (1)
We get,
By taking L.C.M on the R.H.S
We get,
Therefore
Therefore
Now, we know that
Therefore
Now, we know the following trigonometric identity
cos2θ + sin2θ = 1
Therefore,
cos2θ = 1 – sin2θ
Now by substituting the value of sin θ from equation (3)
We get,
Therefore, by taking L.C.M on R.H.S
We get,
Now, by taking square root on both sides
We get,
Therefore,
Substituting the value of sin θ and cos θ from equation (3) and (4) respectively in the equation below
Therefore,
Therefore,
Question: 19
If cos θ = 3/5, find the value of
Solution:
To find the value of
Now we know the following trigonometric identity
cos2θ + sin2θ = 1
Therefore by substituting the value of cos θ from equation (1)
We get,
Therefore,
Therefore by taking square root on both sides
We get,
Now, we know that
Therefore by substituting the value of sin θ and cos θ from equation (2) and (1) respectively
We get,
Now, by substituting the value of sin θ and of tan θ from equation (2) and equation (4) respectively in the expression below
We get,
Therefore,
Question: 20
If sin θ = 3/5, find the value of
Solution:
Given:
Now, we know the following trigonometric identity
cos2θ + sin2θ = 1
Therefore by substituting the value of cos θ from equation (1)
We get,
Therefore,
Now by taking L.C.M
We get,
Therefore, by taking square roots on both sides
We get,
cos θ = 4/5
Therefore,
cos θ = 4/5 … (2)
Now we know that
tan θ = sin θ/cos θ
Therefore by substituting the value of sin θ and cos θ from equation (1) and (2) respectively
We get,
Also, we know that
Therefore from equation (3)
We get,
Now by substituting the value of cos θ, tan θ and cot θ from equation (2), (3) and (4) respectively from the expression below
Therefore,
Question: 21
If tan θ = 24/7, find that sin θ + cos θ
Solution:
Given: tan θ =24/7 … (1)
To find, Sin θ + cos θ
Now we know that tan θ is defined as follows
Now by comparing equation (1) and (2)
We get,
Perpendicular side opposite to ∠θ = 24
Base side adjacent to ∠θ = 7
Therefore triangle representing ∠θ is as shown figure.
Side AC is unknown and can be found by using Pythagoras theorem
Therefore,
AC2 = AB2 + BC2
Now by substituting the value of unknown sides from figure
We get,
AC2= 242 +72
AC = 576 + 49
AC = 625
Now by taking square root on both sides,
We get,
AC = 25
Therefore,
Hypotenuse side AC = 25 …. (3)
Now we know sin θ is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we know that cos θ is defined as follows
Therefore by substituting the value of sin θ and cos θ from equation (4) and (5) respectively, we get
Question: 21
If sin θ = a/b, find sec θ + tan θ in terms of a and b.
Solution:
Given:
sin θ = a/b … (1)
To find: sec θ + tan θ
Now we know, sin θ is defined as follows
Now by comparing equation (1) and (2)
We get,
Perpendicular side opposite to ∠θ = a
Hypotenuse = b
Therefore triangle representing ∠θ is as shown figure.
Hence side BC is unknown
Now we find BC by applying Pythagoras theorem to right angled ΔABC
Therefore,
AC2 = AB2 + BC2
Now by substituting the value of sides AB and AC from figure (a)
We get,
b2 = a2 + BC2
Therefore,
BC2 = b2 – a2
Now by taking square root on both sides
We get,
Therefore,
Now we know cos θ is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we know,
Therefore,
Now we know,
Now by substituting the values from equation (1) and (3)
We get,
Therefore,
Now we need to find sec θ + tan θ
Now by substituting the values of sec θ and tan θ from equation (5) and (6) respectively
We get,
We get,
Now by substituting the value in above expression
We get,
Now,
present in the numerator as well as denominator of above denominator of above expression gets cancels we get,
Square root is present in the numerator as well as denominator of above expression. Therefore we can place both numerator and denominator under a common square root sign
Therefore,
Question: 23
If 8 tan A = 15, find sin A – cos A
Solution:
Given:
8 tan A = 15
Therefore,
To find: sin A – cos A
Now we know tan A is defined as follows
Now by comparing equation (1) and (2)
We get
Perpendicular side opposite to ∠A = 15
Base side adjacent to ∠A = 8
Therefore triangle representing angle A is as shown figure.
Side AC = is unknown and can be found by using Pythagoras theorem
Therefore,
AC2 = AB2 + BC2
Now by substituting the value of known sides from figure (a)
We get,
AC2 = 152 + 82
AC2 = 225 + 64
AC = 289
Now by taking square root on both sides
We get,
AC = √289
AC = 17
Therefore Hypotenuse side AC = 17 … (3)
Now we know, sin A is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we know, cos A is defined as follows
Therefore from figure (a) and equation (3)
We get,
Now we find the value of expression sin A – cos A
Therefore by substituting the value the value of sin A and cos A from equation (4) and (5) respectively, we get,
Question: 24
If tan θ = 20/21, show that
Solution:
Given: tan θ = 20/21
Now we know that
Therefore,
tan θ = 20/21
Side AC be the hypotenuse and can be found by applying Pythagoras theorem
Therefore,
AC2 = AB2 + BC2
AC2= 212 + 202
AC2 = 441 + 400
AC2 = 841
Now by taking square root on both sides
We get,
AC = √841
AC = 29
Therefore Hypotenuse side AC = 29
Now we know, sin θ is defined as follows,
Therefore from figure and above equation
We get,
Now we know cos θ is defined as follows
Therefore from figure and above equation
We get,
Now we need to find the value of expression
Therefore by substituting the value of sin θ and cos θ from above equations, we get
Therefore after evaluating we get,
Hence,
Question: 25
If cosec A = 2, find
Solution:
Given: cosec A = 2
Here BC is the adjacent side,
By applying Pythagoras theorem,
AC2 = AB2 + BC2
4 = 1 + BC2
BC2 = 3
BC = √3
Now we know that
Substitute all the values of sin A, cos A and tan A from the equations (1), (2) and (3) respectively
We get.
= 2
Hence,
Question: 26
If ∠A and ∠B are acute angles such that cos A =cos B, then show that ∠A = ∠B
Solution:
Given:
∠A and ∠B are acute angles
cos A = cos B such that ∠A = ∠B
Let us consider right angled triangle ACB
Now since cos A = cos B
Therefore
Now observe that denominator of above equality is same that is AB
Therefore AC = BC
We know that when two sides of triangle are equal, then opposite of the sides are also
Equal.
Therefore
We can say that
Angle opposite to side AC = angle opposite to side BC
Therefore,
∠B = ∠A
Hence, ∠A = ∠B
Question: 27
In a ΔABC, right angled triangle at A, if tan C = √3, find the value of sin B cos C + cos B sin C.
Solution:
Given: ΔABC
To find: sin B cos C + cos B sin C
The given a ΔABC is as shown in figure
Side BC is unknown and can be found using Pythagoras theorem,
Therefore,
BC2 = AB2 + AC2
BC2 = 3 +1
BC2 = 4
Now by taking square root on both sides
We get,
BC = √4
BC = 2
Therefore Hypotenuse side BC = 2 … (1)
Now,
Therefore,
Now by substituting the values from equation (1) and figure
We get,
Now,
Therefore,
Now substituting the value from equation
Similarly
Now by definition,
So by evaluating
Now, by substituting the value of sin B, cos B, sin C and cos C from equation (2), (3), (4) and (5) respectively in sin B cos C + cos B sin C
Sin B cos C + cos B sin C
= 1
Hence,
sin B cos C + cos B sin C = 1
Question: 28
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 = 12/5 for some value of ∠A.
(iii) cos A is the abbreviation used for the cosecant of ∠A.
(iv) sin θ = 4/3 for some angle θ.
Solution:
(i) tan A < 1
Value of tan A at 45° i.e… tan 45 = 1
As value as A increases to 90°
Tan A becomes infinite
So given statement is false.
(ii) sec A = 12/5 for some value of angle if
M-I
sec A = 2.4
sec A > 1
So given statements is true.
M- II
For sec A = 12/5 we get adjacent side = 13
Subtending 9i at B.
So, given statement is true.
(iii) Cos A is the abbreviation used for cosecant of angle A.
The given statement is false.
As such cos A is the abbreviation used for cos of angle A, not as cosecant of angle A.
(iv) cot A is the product of cot A and A
Given statement is false? cot A is a co-tangent of angle A and co-tangent of angle
(v) sin θ = 4/3 for some angle θ.
Given statement is false
Since value of sin θ is less than (or) equal to one.
Here value of sin θ exceeds one,
So given statement is false.
Question: 29
Solution:
As shown in figure
Here BC is the adjacent side,
By applying Pythagoras theorem,
AC2 = AB2 + BC2
169 = 144 + BC2
BC2 = 169 – 144
BC2 = 25
BC = 5
Now we know that,
We also know that,
tan θ = sin θ/cos θ
Therefore, substituting the value of sin θ and cos θ from above equations
We get,
tan θ = 12/5
Now substitute all the values of sin θ, cos θ and tan θ from above equations in
We get,
Therefore by further simplifying we get,
Therefore,
Hence,
Question: 30
If cos θ = 5/13, find the value of
Solution:
Given: If cos θ = 5/13 ….. (1)
To find: The value of expression
Now we know that
Now when we compare equation (1) and (2)
We get,
Base side adjacent to ∠θ = 5
Hypotenuse = 13
Therefore, Triangle representing ∠θ is as shown figure.
Perpendicular side AB is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
AC2= AB2 + BC2
Therefore by substituting the values of known sides,
AB2 = 132 – 52
AB2= 169 -25
AB2 = 144
AB = 12 …. (3)
Now we know from figure and equation,
Now we know that,
Now w substitute all the values from equation (1), (4) and (5) in the expression below,
Therefore
We get,
Therefore by further simplifying we get,
Therefore,
Hence,
Question: 31
Solution:
Given: sec A = 17/8
Now we know that
Now, by substituting the value of sec A
We get,
Now we also know that,
sin2A + cos2A = 1
Therefore
sin2A = 1 – cos2A
Now by taking square root on both sides,
We get,
sin A = 15/17
We also know that,
Now by substituting the value of all the terms,
We get,
tan A = 15/8
Now from the expression of above equation which we want to prove:
Now by substituting the value of cos A ad sin A from equation (3) and (4)
We get,
From expression
Now by substituting the value of tan A from above equation
We get,
Therefore,
We can see that,
Question: 32
If sin θ = 3/4,
Prove that
Solution:
To prove:
By definition,
By comparing (1) and (3)
We get,
Perpendicular side = 3 and
Hypotenuse = 4
Side BC is unknown.
So we find BC by applying Pythagoras theorem to right angled ΔABC
Hence,
AC2 = AB2 + BC2
Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)
Therefore,
42 = 32 + BC2
BC2 = 16 – 9
BC2 = 7
BC = √7
Hence, Base side BC = √7 … (3)
Now cos A = BC/AC
√7/4 … (4)
Now, cosec A = 1/(sin A)
Therefore, from fig and equation (1)
cosec A = 4/3 … (5)
Now, similarly
Sec A = 4/√7 … (6)
Further we also know that
Therefore by substituting the values from equation (1) and (4),
We get,
Now by substituting the value of cosec A, sec A and cot A from the equations (5), (6), and (7) in the L.H.S of expression (2)
Hence it is proved that,
Question: 33
If sec A = 17/8,
Verify that
Solution:
Given: sec A = 17/8 … (1)
To verify:
Now we know that
sec A = 1/(cos A)
Therefore,
cos A = 1/(sec A)
We get,
cos A = 8/17 … (3)
Similarly we can also get,
sin A = 15/17 …. (4)
An also we know that
tan A = (sin A)/(cos A)
tan A = 15/8 … (5)
Now from the expression of equation (2)
L.H.S: Missing close
Now by substituting the value of cos A and sin A from equation (3) and (4)
We get,
Now by substituting the value of tan A from equation (5)
We get,
Now by comparing equation (6) and (7)
We get,
Question: 34
If cot θ = 3/4, prove that
Solution:
Given: cot θ = 3/4
Prove that:
Now we know that
Here AC is the hypotenuse and we can find that by applying Pythagoras theorem
AC2 = AB2 +BC2
AC2 = 16 +9
AC2 = 25
AC = 5
Similarly
Now on substituting the values in equations we get,
Therefore,
Question: 35
If 3 cos θ – 4 sin θ = 2 cos θ + sin θ, find tan θ tan θ
Solution:
Given: 3 cos θ – 4 sin θ = 2 cos θ + sin θ
To find: tan θ tan θ
We can write this as:
3 cos θ – 4 sin θ = 2 cos θ + sin θ cos θ = 5 sin θ
Dividing both the sides by cos θ,
We get,
1 = 5 tan θ
tan θ = 1
Hence,
tan θ = 1
Question: 36
If ∠A and ∠P are acute angles such that tan A = tan P, then show ∠A = ∠P
Solution:
Given: A and P are acute angles tan A = tan P
Prove that: ∠A = ∠P
Let us consider right angled triangle ACP
We know
∴ tan A = tan P
PC = AC [∵ Angle opposite to equal sides are equal]
∠A = ∠P
All Chapter RD Sharma Solutions For Class10 Maths
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