NCERT solutions for class 10 Maths
Chapter 8
Coordinate Geometry
Exercise 8.1
Ex 8.1 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 a ΔABC,
Using Pythagoras Theorem,
AC2=AB2+BC2
AC2 = (24)2+(7)2
AC2 = (576+49)
AC2 = 625cm2
AC = √625 = 25
∴AC = 25 cm
(i) Sin (A), Cos (A)
Sin (A) = BC/AC = 7/25
Cos (A) = AB/AC = 24/25
(ii) Sin (C), Cos (C)
Sin (C) = AB/AC = 24/25
Cos (C) = BC/AC = 7/25
Ex 8.1 Question 2.
In Fig. 8.13, find tan P – cot R
Solution:
In Δ PQR,
PR = 13cm,
PQ = 12cm
Using Pythagoras Theorem,
PR2 = QR2 + PQ2
132 = QR2+122
169 = QR2+144
QR2 = 169−144
QR2 = 25
QR = √25 = 5
tan (P) – cot (R) = QR/PQ – QR/PQ = 5/12 – 5/12 = 0
∴tan(P) – cot(R) = 0
Ex 8.1 Question 3.
If sin A = 3/4, Calculate cos A and tan A.
Solution:
Given: Sin A = 3/4
Let, Sin A = 3k/4k [where k is a positive real number.]
In ΔABC
Using Pythagoras theorem,
AC2=AB2 + BC2
(4k)2=AB2 + (3k)2
16k2−9k2 =AB2
AB2=7k2
∴AB = √7k
Cos (A) = AB/AC = √7k/4k = √7/4
∴ cos (A) = √7/4
tan(A) = BC/AB = 3k/√7k = 3/√7
∴ tan A = 3/√7
Ex 8.1 Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
Given: 15 cot A = 8
So, Cot A = 8/15
Let, cot A = 8k/15k [Where, k is a positive real number.]
In ΔABC
Using Pythagoras theorem,
AC2=AB2 + BC2
AC2= (8k)2 + (15k)2
AC2= 64k2 + 225k2
AC2= 289k2
∴AC = 17k
Sin (A) = BC/AC = 15k/17k = 15/17
∴sin A = 15/17
Sec (A) =AC/AB = 17k/8k = 17/8
∴ sec (A) = 17/8
Ex 8.1 Question 5.
Given sec θ = 13/12 Calculate all other trigonometric ratios
Solution:
sec θ =13/12
Let sec θ =13k/12k [Where, k is a positive real number.]
In ΔABC
Using Pythagoras theorem,
AC2=AB2 + BC2
(13k)2= (12k)2 + BC2
169k2= 144k2 + BC2
169k2= 144k2 + BC2
BC2 = 169k2 – 144k2
BC2= 25k2
∴ BC = 5k
Sin θ = BC/AC = 5/13
Cos θ = AB/AC = 12/13
tan θ = BC/AB = 5/12
Cosec θ = AC/BC = 13/5
cot θ = AB/BC = 12/5
Ex 8.1 Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.
Solution:
In ΔABC
Using Pythagoras theorem,
Given ,Cos (A) = cos (B)
Cos (A) = cos (B)
⇒ AP/PQ = BC/BD
⇒ AP/BC = AQ/BD
Ex 8.1 Question 7.
If cot θ = 7/8, evaluate :
(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2 θ
Solution:
Given: cot θ = 7/8
Let, cot θ = 7k/8k [where k is a positive real number]
In△ABC
Using Pythagoras theorem
AC2 = AB2+BC2
AC2 = (7k)2+(8k)2
AC2 = 49k2+64k2
AC2 = 113k2
AC = √113 k
Ex 8.1 Question 8.
If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A or not.
Solution:
Given: 3 cot A = 4
cot(A) =4/3
Let, cot(A) = 4k/3k [where k is a positive real number.]
In△ABC
Using Pythagorean theorem,
AC2=AB2+BC2
AC2=(4k)2+(3k)2
AC2=16k2+9k2
AC2=25k2
AC=5k
tan(A) = BC/AB = 3/4
sin (A) = BC/AC = 3/5
cos (A) = AB/AC = 4/5
Ex 8.1 Question 9.
In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
Given: tan A = 1/√3
Let tan A = 1k/√3k [Where k is the positive real number]
In ΔABC
Using Pythagorean theorem,
AC2=AB2+BC2
AC2=(√3 k)2+(k)2
AC2=3k2+k2
AC2=4k2
AC = 2k
Sin A = 1/2
Cos A = √3/2
Ex 8.1 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:
Ex 8.1 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 = 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 = 4/3 for some angle θ.
Solution:
(i) False
Because, tan A = opposite side of angle A / Adjecent side of angle A , If opposite side >Adjecent side, then the value of tan A is greater than 1.
(ii) True
Because, sec A = Hypotenuse / Adjecent side of angle A.Then hypotenuse is always greater than adjecent side.
(iii) False
Because, cos A is used for cosine of angle A.
(iv) False
Because, cot A is used for cotangent of angle A.
(v) False
Because,sin θ = opposite side of angle A /Hypotenuse. Hypotenuse is always greater than opposite side.