﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 8.1

Question-1 :-  In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.
Determine : (i) sin A, cos A  (ii) sin C, cos C

Solution :-
```  Given :
In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm, AC = ?
By Pythagoras Theorem :
AC² = AB² + BC²
= (24)² + (7)²
= 576 + 49
= 625
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
```

Question-2 :-  In Figure, find tan P – cot R. Solution :-
```  Given :
In Δ PQR, right-angled at Q, PQ = 12 cm, PR = 13 cm, QR = ?
By Pythagoras Theorem :
QR² = PR² - PQ²
= (13)² - (12)²
= 169 - 144
= 25
QR = 5 cm

tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
tan P - cot R = 5/12 - 5/12 = 0
```

Question-3 :-  If sin A = 3/4, calculate cos A and tan A.

Solution :-
```  Given :
In Δ ABC, right-angled at B,
sin A = 3/4 = BC/AC
AC = 4 cm, BC = 3 cm, AB = ?
By Pythagoras Theorem :
AB² = AC² - BC²
= (4)² - (3)²
= 16 - 9
= 7
AB = √7 cm cos A = AB/AC = √7/4
tan A = BC/AB = 3/√7
```

Question-4 :-  Given 15 cot A = 8, find sin A and sec A.

Solution :-
```  Given :
In Δ ABC, right-angled at B,
15 cot A = 8
cot A = 8/15 = AB/BC
AB = 8 cm, BC = 15 cm, AC = ?
By Pythagoras Theorem :
AC² = AB² + BC²
= (8)² + (15)²
= 64 + 225
= 289
AC = 17 cm sin A = BC/AC = 15/17
sec A = AC/AB = 17/8
```

Question-5 :-  Given sec θ = 13/12 calculate all other trigonometric ratios.

Solution :-
```  Given :
In Δ ABC, right-angled at B,
sec θ = 13/12 = AC/AB
AB = 12 cm, BC = ?, AC = 13 cm
By Pythagoras Theorem :
BC² = AC² - AB²
= (13)² - (12)²
= 169 - 144
= 25
BC = 5 cm sin A = BC/AC = 5/13
cos A = AB/AC = 12/13
tan A = BC/AB = 5/12
cot A = AB/BC = 12/5
cosec A = AC/BC = 13/5
```

Question-6 :-  If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution :-
```  Let us consider CD Perpendicular to AB Given :
cos A = cos B
Let AD/BD = AC/BC = k
AC = k BC .....(ii)
By using Pythagoras Theorem for Δ CAD and Δ CBD, we obtain :
CD² = AC² - AD² .....(iii)
CD² = BC² - BD² .....(iv)
from equation (iii) & (iv), we obtain
AC² - AD² = BC² - BD²
(k BC)² - (k BD)² = BC² - BD²
k² BC² - k² BD² = BC² - BD²
k²(BC² - BD²) = BC² - BD²
k² = 1
k = ±1
K = 1
Putting this value in equation (ii), we obtain
AC = BC
So, ∠ A = ∠ B (Angles opposite to equal sides of a triangle)
```

Question-7 :-  If cot θ = 7/8, evaluate :
(i) (ii) cot² θ

Solution :-
```  Given :
In Δ ABC, right-angled at B,
cot θ = 7/8 = AB/BC
AB = 7 cm, BC = 8 cm, AC = ?
By Pythagoras Theorem :
AC² = AB² + BC²
= (7)² + (8)²
= 49 + 64
= 113
AC = √113 cm sin θ = BC/AC = 8/√113
cos θ = AB/Ac = 7/√113
(i) (ii) cot² θ = (cos θ/sin θ)²
= (7/√113 x √113/8)²
= (7/8)²
= 49/64
```

Question-8 :-  If 3 cot A = 4, check whether or Not.

Solution :-
```  Given :
In Δ ABC, right-angled at B,
3 cot A = 4
cot A = 4/3 = AB/BC
AB = 4 cm, BC = 3 cm, AC = ?
By Pythagoras Theorem :
AC² = AB² + BC²
= (4)² + (3)²
= 16 + 9
= 25
AC = 5 cm sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
tan A = BC/AB = 3/4

Now, L.H.S R.H.S
cos² A - sin² A = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25
L.H.S = R.H.S
```

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 :
In Δ ABC, right-angled at B,
tan A = 1/√3 = BC/AB
AB = √3 cm, BC = 1 cm, AC = ?
By Pythagoras Theorem :
AC² = AB² + BC²
= (√3)² + (1)²
= 3 + 1
= 4
AC = 2 cm sin A = BC/AC = 1/2
cos A = AB/AC = √3/2
sin C = AB/AC = √3/2
cos C = BC/AC = 1/2

(i) sin A cos C + cos A sin C
= 1/2 x 1/2 + √3/2 x √3/2
= 1/4 + 3/4
= 4/4
= 1

(ii) cos A cos C – sin A sin C
= √3/2 x 1/2 - 1/2 x √3/2
= √3/4 - √3/4
= 0
```

Question-10 :-  In Δ 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 :-
```  Given :
In Δ PQR, right-angled at Q,
PR + QR = 25 cm and PQ = 5 cm
PR = 25 - QR
By Pythagoras Theorem :
PR² = PQ² + QR²
(25 - QR)² = (5)² + QR²
25² + QR² - 50 QR = 25 + QR²
625 - 50 QR = 25
-50 QR = 25 - 625
-50 QR = -600
QR = 600/50
QR = 12 cm
Now, PR = 25 - QR = 25 - 12 = 13 cm sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
```

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) Consider a Δ ABC, right-angled at B. tan A = 10/5
But 10/5 > 1
tan A > 1
So, tan A < 1 in not always true.
Hence, the given statement is false.
```
```(ii) sec A = 12/5
AC/AB = 12/5
In Δ ABC, right-angled at B,
AC = 12 cm, AB = 5 cm, BC = ?
By Pythagoras Theorem :
BC² = AC² - AB²
= (12)² - (5)²
= 144 - 25
= 119
BC = 10.9 It can be observe that for give two sides AC = 12 and AB = 5
BC should not be such that,
(AC - AB) < BC < (AC + AB)
(12 - 5) < BC < (12 + 5)
7 < BC < 17
However, BC = 10.9. Clearly, such a triangle is possible and hence,
such value of sec A is possible.
Hence the given statement is true.
```
```(iii) Abbreviation used for cosecant of angle A is cosec A and cos A is the
abbreviation used for cosine of angle A.
Hence, the given statement is false.
```
```(iv) cot A is not a product of cot and A. It is the cotangent of angle A.
Hence, the given statement is false.
```
```(v) sin θ = 4/3
We know that in a Right angled Triangle.
sin θ = Perpendicular/Hypotaneouss
In a right angled triangle, hypotaneous is always greater than the remaining two sides.
Therefore, such value of sin θ is not possible.
Hence, the given statement is false.
```
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