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SEBA Class 10 Mathematics Chapter 8 Introduction to Trigonometry
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Introduction to Trigonometry
Chapter – 8
| Exercise 8.1 |
1. In ∆ABC right angled at B, AB = 24cm, BC = 7cm. Determine:
(i) sin A, cosA
(ii) sin C, cosC
Ans: Applying Pythagoras theorem for ∆ABC, we obtain
AC2 = AB2 + BC2
= (24 cm)2 + (7 cm)2
= (576 + 49) cm2
= 625 cm2
∴ AC = cm = 25 cm
2. In the given figure find tanP − cotR.
Ans: Applying Pythagoras theorem for ∆PQR, we obtain
PR2 = PQ2 + QR2
(13 cm)2 = (12 cm)2 + QR2
169 cm2 = 144 cm2 + QR2
25 cm2 = QR2
QR = 5 cm
3. If sin A = 3/4, calculate cosA and tanA.
Ans:
4. Given 15 cotA = 8, find sinA and secA.
Ans: Consider a right-angled triangle, right-angled at B.
5. Given secθ = 13/12, calculate all other trigonometric ratios.
Ans:
If AC is 13k, AB will be 12k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
(AC)2 = (AB)2 + (BC)2
(13k)2 = (12k)2 + (BC)2
169k2 = 144k2 + BC2
25k2 = BC2
BC = 5k
6. If ∠A and ∠B are acute angles such that cosA = cosB, then show that ∠A = ∠B.
Ans: Let us consider a triangle ABC in which CD ⊥ AB.
It is given that cosA = cosB
AB/AC = BD/BC …(1)
We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.
7. If cot θ = ⅞ evaluate
Ans: Let us consider a right triangle ABC, right-angled at point B.
If BC is 7k, then AB will be 8k, where k is a positive integer. Applying Pythagoras theorem in ∆ABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (7k)2
= 64k2 + 49k2
= 113k2
(ii) cot2 θ
Ans:
Ans: If 3 cot A = 4,
It is given that 3 cot = 3/4 A = 4 Or, cot A = Consider a right triangle ABC, right-angled at point B.
If AB is 4k, then BC will be 3k, where k is a positive integer.
In ∆ABC,
(AC)2 = (AB)2 + (BC)2
= (4k)2 + (3k)2
= 16k2 + 9k2
= 25k2
AC = 5k
9. In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:
(i) sinA cosC + cosA sinC
(ii) cosA cosC – sinA sinC
10. In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.
Ans: Given that, PR + QR = 25 cm
PQ = 5 cm
Let PR be x.
Therefore, QR = (25 − x)cm
Applying Pythagoras theorem in ∆PQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13
Therefore, PR = 13 cm
QR = (25 − 13) cm = 12 cm
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
Ans: (i) Consider a ∆ABC, right-angled at B.
So, tan A < 1 is not always true.
Hence, the given statement is false.
Let AC be 12k, AB will be 5k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
144k2 = 25k2 + BC2
BC2 = 119k2.
BC = 10.9k
It can be observed that for given two sides AC = 12k and AB = 5k,
BC should be such that,
AC − AB < BC < AC + AB
12k − 5k < BC < 12k + 5k
7k < BC < 17 k
However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible.
Hence, the given statement is true.
(iii) cos A is the abbreviation used for the cosecant of angle A.
Ans: The abbreviation used for cosecant of angle A is cosec A. And cos A is the
abbreviation used for the cosine of angle A.
Hence, the given statement is false.
(iv) cot A is the product of cot and A
Ans: cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false.
(v) sin θ = 3/4, for some angle θ
Ans: sin θ = ¾ We know that in a right-angled triangle,
In a right-angled triangle, the hypotenuse is always greater than the remaining two sides.
Therefore, such a value of sin θ is not possible.
Hence, the given statement is false.
| Exercise 8.2 |
1. Evaluate the following:
(i) sin60° cos30° + sin30° cos 60°
Ans: sin60° cos30° + sin30° cos 60°
(ii) 2tan245° + cos230° − sin260°
Ans:
Ans:
2. Choose the correct option and justify your choice.
(a) sin60°
(b) cos60°
(c) tan60°
(d) sin30°
Ans: (a) sin60°
(a) tan90°
(b) 1
(c) sin45°
(d) 0
Ans: (d) 0
(iii) sin2A = 2sinA is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Ans: (a) 0°
(a) cos60°
(b) sin60°
(c) tan60°
(d) sin30°
Ans: (c) tan60°
3.
Ans: tan (A + b) √3
tan (A + B) = tan 60
⇒ A + B = 60 ….(1)
⇒ A – B = 30 …(2)
On adding both equations, we obtain
2A = 90
⇒ A = 45
From equation (1), we obtain
45 + B = 60
B = 15
Therefore, ∠A = 45° and ∠B = 15°
4. Value of sinθ cos(90° – θ) + cosθ sin(90° – θ) is-
(a) -1
(b) 0
(c) 1
(d) 2
Ans: (c) 1
5. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B
Ans: sin (A + B) = sin A + sin B Let A = 30° and B = 60°
sin (A + B) = sin (30° + 60°)
= sin 90° = 1
And sin A + sin B = sin 30° + sin 60°
Clearly, sin (A + B) ≠ sin A + sin B
Hence, the given statement is false.
(ii) The value of sinθ increases as θ increases.
Ans: The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as sin 0° = 0
(iii) The value of cos θ increases as θ increases
Ans:
cos 90° = 0
It can be observed that the value of cos θ does not increase in the interval of 0°<θ<90°.
Hence, the given statement is false.
(iv) sinθ = cos θ for all values of θ
Ans: sin θ = cos θ for all values of θ. This is true when θ = 45°
It is not true for all other values of θ.
(v) cot A is not defined for A = 0°
Ans:
| Exercise 8.3 |
1. Evaluate:
Ans:
Ans:
Ans:
(iv) cosec 31° − sec 59°
= cosec (90° − 59°) − sec 59°
= sec 59° − sec 59°
= 0
2. Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
Ans: tan 48° tan 23° tan 42° tan 67°
= tan (90° − 42°) tan (90° − 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
= 1
(ii) cos 38° cos 52° − sin 38° sin 52° = 0
Ans: cos 38° cos 52° − sin 38° sin 52°
= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°
= sin 52° sin 38° − sin 38° sin 52°
= 0
3. If tan 2A = cot (A − 18°), where 2A is an acute angle, find the value of A.
Ans: Given that, tan 2A = cot (A− 18°)
⇒ cot (90° − 2A) = cot (A −18°)
⇒ 90° − 2A = A− 18°
⇒ 108° = 3A
A = 36°
4. If tan A = cot B, prove that A + B = 90°
Ans: Given that, tan A = cot B
tan A = tan (90° − B)
A = 90° − B
A + B = 90°
5. If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Ans: Given that, sec 4A = cosec (A − 20°)
⇒ cosec (90° − 4A) = cosec (A − 20°)
⇒ 90° − 4A = A− 20°
⇒ 110° = 5A
A = 22°
6. If A, B and C are interior angles of a triangle ABC, then show that
Ans: We know that for a triangle ABC,
∠ A + ∠B + ∠C = 180°
∠B + ∠C= 180° − ∠A
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Ans: sin 67° + cos 75°
= sin (90° − 23°) + cos (90° − 15°)
= cos 23° + sin 15°
8. (i) If sec5θ = cosec(θ – 36°) where θ is an acute angle. Then find the value of θ.
Ans: Sec5θ = Cosec (θ – 36°)
Again, Sec5θ = Cosec (90° – 5θ)
∴ Cosec (90° – 5θ) = Cosec (θ – 36°)
⇒ 90° – 5θ = θ – 36°
⇒ -6θ = -126° ∴ θ = 126°/6 = 21°
(ii) If sinA = cos 30°, A < 90°. Find the value of A.
Ans: Sin A = Cos 33°
Sin A = Cos(90° – A)
∴ 90° – A = 33°
⇒ -A = 33° – 90°
⇒ -A = -57 ∴ A = 57°
(iii) sin2A = cos(A + 15°) where 2A < 90°. Find the value of A.
Ans: Sin2 A = Cos (A + 15°)
Sin2 A = Cos (90° – 2A)
∴ Cos (A + 15°) = Cos (90° – 2A)
⇒ A + 15° = 90° – 2A
⇒ 3A = 90° – 15°
⇒ 3A = 75° ∴ A = 75°/3 = 25°
(iv) If sin(3x + 10) = cos(x + 24), then find the value of x.
Ans: Sin (3x + 10°) = Cos (x + 24°)
Sin (3x + 10°) = Cos {90° – (3x + 10°)}
∴ Cos (x + 24°) = Cos {90° – (3x + 10°)}
⇒ x + 24° = 90° – 3x – 10°
⇒ 4x = 80° – 24°
⇒ 4x = 56°
∴ x = 56°/4 = 14°
| Exercise 8.4 |
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Ans:
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Ans: We know that,
3. Evaluate:
Ans:
Ans:
4. Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =
(a) 1
(b) 9
(c) 8
(d) 0
Ans: (b) 9
(ii) (1 + tan + sec ) (1 + cot – cosec ) =
(a) 0
(b) 1
(c) 2
(d) -1
Ans: (a) 0
(iii) (sec A + tan A) (1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Ans: (d) cos A
(iv) 1 + tan2 A/1 + Cot2 A
(a) sec2 A
(b) -1
(c) cot2 A
(d) tan2 A
Ans: (d) tan2 A.
54. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
Ans:
Ans:
Ans:
Ans:
Ans:
Ans:
Ans:
6. Prove that:
Ans: LHS = tan4θ + tan2θ
= tan2θ + (1 + tan2θ)
= (Sec2θ – 1). Sec2θ
= Sec4θ – Sec2θ = RHS.
Ans:
Ans:
(iv) cotθ + tanθ = secθ = secθcosecθ
Ans: LHS = Cotθ + tanθ
Ans:
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