Complete NCERT solutions for class 10 Maths Chapter 8 Introduction To Trigonometry

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Written by Team Trustudies
Updated at 2021-02-16


Solution for Exercise -8.1

Q1. In ? ABC , right angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A
(ii) sin C, cos C



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given, in ? ABC,
AB = 24 cm, BC = 7 cm and right angled at B



By using the Pythagoras theorem, we have

AC2 = AB2 + BC2
=> AC2 = (24)2 + (7)2
=> AC2 = 576 + 49 = 625

=> ? AC = 625 = 25 cm

(i) sinA = BCAC = 725 ???? [? sin? = PH ]

cosA = ABAC = 2425 ???? [? cos? = BH ]

(ii) sinC = ABAC = 2425

cosC = BCAC = 725

Q2. In Fig. find tan P – cot R.




NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


By using the Pythagoras theorem, we have

PR2 = PQ2 + QR2

=> 132 = 122 + QR2

=> QR2 = 132 ? 122 = 169 ? 144 = 25

=> QR = 25 = 5

? tanP = PB = QRPQ = 512
cotR = BP = QRPQ = 512

Hence, tanP  cotR = 512 ? 512 = 0

Q3. If sinA = 34 , calculate cos A and tan A.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given,
sinA = 34 = PH = BCCA

Let BC = 3k and AC = 4k.



Then, AB = AC2?BC2 = (4k)2?(3k)2 = 16k2?9k2 = 7k2 = 7k

? cosA = BH = ABAC = 7k4k = 74

tanA = PB = BCAB = 3k7k = 37

Q4. Given 15 cot A = 8, find sin A and sec A.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given,
15cotA = 8
=> cotA = 815 = BP = ABBC

Let AB = 8k and BC = 15k.

Then, by Pythagoras theorem

AC = AB2+BC2 = (8k)2+(15k)2 = 64k2+225k2 = 289k2 = 17k



? sinA = PH = BCAC = 15k17k = 1517

and, secA = HB = ACAB = 17k8k = 178

Q5. Given sec? = 1312 , calculate all other trigonometric ratios.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given, sec? = 1312 = HB = ACAB

Let AC = 13k and AB = 12k.

Then by Pythagoras theorem,

BC = AC2?AB2 = (13k)2?(12k)2 = 169k2?144k2 = 25k2 = 5k



? sin? = PH = BCAC = 5k13k = 513

cos? = BH = ABAC = 12k13k = 1213

tan? = PB = BCAB = 512

cot? = 1tan? = 125

cosec? = 1sin? = 135

Q6. If ? A and ? B are acute angles such that cos A = cos B, then show that ? A = ? B.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :




In ? ABC,
cosA = ACAB = BH

and, cosB = BCAB

But given that, cos A = cos B

=> ACAB = BCAB
=> AC = BC

? ? A = ? B [? Angles opposite to equal sides are equal]

Q7 If cot? = 78 , evaluate :

(i) (1+sin?)(1?sin?)(1+cos?)(1?cos?)

(ii) cot2?



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given, cot? = BP = ABBC = 78

Let AB = 7k and BC = 8k.
Then by Pythagoras theorem,

AC = BC2+AB2 = (8k)2+(7k)2 = 64k2+49k2 = 113k2 = 113k



? sin? = PH = BCAC = 8k113k = 8113

and, cos? = BH = ABAC = 7k113k = 7113

(i) ? (1+sin?)(1?sin?)(1+cos?)(1?cos?) = 1?sin2?1?cos2?

= 1?641131?49113

= 113?64113?49

= 4964

(ii) cot2? = (78)2 = 4964

Q8. If 3cotA = 4, check whether 1?tan2A1+tan2A = cos2A ? sin2A or not.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given, 3cotA = 4 = 43 = BP = ABBC

Let AB = 4k and BC = 3k

Then by Pythagoras theorem,

AC = AB2+BC2 = (4k)2+(3k)2 = 16k2+9k2 = 25k2 = 5k



? sinA = PH = BCAC = 3k5k = 35

cosA = BH = ABAC = 4k5k = 45

and, tanA = 1cotA = 34

Now, L.H.S. = 1?tan2A1+tan2A = 1?9161+916 = 16?916+9 = 725

R.H.S. = cos2A ? sin2A = (45)2 ? (35)2 = 1625 ? 925 = 725

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

? 1?tan2A1+tan2A = cos2A ? sin2A

Q9. In ? ABC right angled at B, if tanA = 13 , find the value of

(i) sinAcosC + cosAsinC
(ii) cosAcosC?sinAsinC



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given, tanA = PH = BCAB = 13

Let BC = k and AB = 3k.

Then by Pythagoras theorem,

AC = AB2+BC2 = 3k2+k2 = 4k2 = 2k



? sinA = PH = BCAC = k2k = 12

cosA = BH = ABAC = 3k2k = 32

sinC = PH = ABAC = 3k2k = 32

and, cosC = BH = BCAC = k2k = 12

(i) sinAcosC + cosAsinC = 12×12 + 32×32 = 14 + 34 = 44 = 1

(ii) cosAcosC ? SinAsinC = 32×12 ? 12×32 = 0

Q10. 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.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


Given, PR + QR = 25 cm and PQ = 5cm

Let QR = x cm
Thus, PR = (25 – x) cm



Then by Pythagoras theorem,

RP2 = RQ2 + QP2

(25?x)2 = x2 + 52

625 ? 50x + x2 = x2 + 25

?50x = ?600

=> x = 60050 = 12

? RQ = 12 cm

=> RP = (25 – 12) = 13 cm

Now, sinP = PH = RQRP = 1213

cosP = BH = PQRP = 513

and tanP = PB = RQPQ = 125

Q11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.

(ii) secA = 125 for some value of angle A .

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

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

(v) sin? = 43 for some angle ?.



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


(i) False because sides of a right triangle may have any length, so tan A may have any value.

(ii) True as sec A is always greater than 1.

As, secA = HB and Hypotenuse is the largest side ? sec A is always greater than 1

(iii) False as cos A is the abbreviation used for cosine A.

(iv) False as cot A is not the product of 'cot' and A. 'cot' separated from A has no meaning.

(v) False as sin ? cannot be > 1

Solution for Exercise -8.2

Q1. Evaluate :
(i) sin60°cos30°+sin30°cos60°

(ii) 2tan245°+cos230°?sin260°

(iii) cos45°sec30°+cosec30°

(iv) sin30°+tan45°?cosec60°sec30°+cos60°+cot45°

(v) 5cos260°+4sec230°?tan245°sin230°+cos230°



NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry


Answer :


(i) sin60°cos30° + sin30°cos60 = 32×32 + 12×12 = 34 + 14 = 3+14 = 1



(ii)2tan245° + cos230° ? sin260° = 2(1)2 + (32)2 ? (32)2 = 2 + 34 ? 34 = 2



(iii) cos45°sec30°+cosec30° = 1223+2 = 122+233 = 12×32+23 = 32×2(3+1)×3?13?1 = 3(3?1)2×2(3?1) = 2×3(3?1)2×2×2×2 = 32?68



(iv) sin30°+tan45°?cosec60°sec30°+cos60°+cot45° = 12+1?2323+12+1 = 1+22?2323+1+22 = 32?2323+32 = 33?4234+3323 = 33?44+33 = (33?4)(33?4)(4+33)(33?4) = 27+16?123?12327?16 = 43?24311



(v) 5cos260°+4sec230°?tan245°sin230°+cos230° = 5(12)2+4(23)2?(1)2(12)2+(32)2 = 5×14+4×43?114+34 = 54 + 163?1 = 15+64?1212 = 6712

Q2. Choose the correct option and justify :

(i) 2tan30°1+tan230° =

(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) 1?tan245°1+tan245° =

(A) tan 90°
(B) 1
(C) sin 45 °
(D) 0

(iii)