**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,

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

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

AC^{2} = (576+49)

AC^{2} = 625cm^{2}

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,

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

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

169 = QR^{2}+144

QR^{2 }= 169−144

QR^{2 }= 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,

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

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

16k^{2}−9k^{2 }=AB^{2}

AB^{2}=7k^{2}

∴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,

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

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

AC^{2}= 64k^{2} + 225k^{2}

AC^{2}= 289k^{2}

∴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,

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

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

169k^{2}= 144k^{2} + BC^{2}

169k^{2}= 144k^{2} + BC^{2}

BC^{2 = }169k^{2} – 144k^{2}

BC^{2}= 25k^{2}

∴ 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) cot ^{2} θ**

**Solution:**

Given: cot θ = 7/8

Let, cot θ = 7k/8k [where k is a positive real number]

In△ABC

Using Pythagoras theorem

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

AC^{2 }= (7k)^{2}+(8k)^{2}

AC^{2 }= 49k^{2}+64k^{2}

AC^{2 }= 113k^{2}

AC = √113 k

**Ex 8.1 Question 8.**

** If 3 cot A = 4, check whether (1-tan ^{2 }A)/(1+tan^{2} A) = cos^{2} 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,

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

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

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

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

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,

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

AC^{2}=(√3 k)^{2}+(k)^{2}

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

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

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.