﻿ Class 10 NCERT Math Solution
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TOPICS
Unit-8(Examples)

Example-1 :-  Given tan A = 4/3, find the other trigonometric ratios of the angle A.

Solution :-
```  Give that:

tan A = 4/3
cot A = 1/tan A = 3/4
sec² A = 1 + tan² A
= 1 + (4/3)²
= 1 + 16/9
= 25/9
sec A = 5/3
cos A = 1/sec A = 3/5
sin² A = 1 - cos² A
= 1 - (3/5)²
= 1 - 9/25
= 16/25
sin A = 4/5
cosec A = 1/sin A = 5/4
```

Example-2 :-  If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.

Solution :-
```  Let us consider two right triangles ABC and PQR where sin B = sin Q see in figure

We have sin B = AC/AB
sin Q = PR/PQ
Then, AC/AB = PR/PQ
Therefore, AC/PR = AB/PQ = k  .......(1)
Now using Pythagoras theorem,

From (1) and (2), we have
AC/PR = AB/PQ = BC/QR
Δ ACB ~ Δ PRQ and therefore, ∠ B = ∠ Q.
```

Example-3 :-  Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Figure). Determine the values of (i) cos² θ + sin² θ,  (ii) cos² θ – sin² θ

Solution :-
```  In Δ ACB, we have AB = 29 units and BC = 21 units
AC² = AB² - BC²
= (29)² - (21)²
= (29 - 21)(29 + 21)
= 8 x 50
= 400
AC = 20 units
so, sin θ = AC/AB = 20/29, cos θ = 21/29
```
```  Now,
(i) cos² θ + sin² θ
= (21/29)² + (20/29)²
= 441/841 + 400/841
= 841/841
= 1
```
```(ii)cos² θ – sin² θ
= (21/29)² - (20/29)²
= 441/841 - 400/841
= 41/841
```

Example-4 :-  In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

Solution :-
```
In Δ ABC, tan A = BC/AB = 1 (see in figure)
i.e., BC = AB
Let AB = BC = k, where k is a positive number
AC² = AB² + BC²
= k² + k²
= 2k²
AC = √2k
Therefore, sin A = BC/AC = k/√2k = 1/√2
cos A = AB/AC = 1/√2
so, 2 sin A . cos A = 2 (1/√2) x (1/√2) = 1
which is the required value.
```

Example-5 :-  In Δ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Figure). Determine the values of sin Q and cos Q.

Solution :-
```  In Δ OPQ, we have
OQ² = OP² + PQ²
(1 + PQ)² = OP² + PQ²
i.e., 1 + PQ² + 2PQ = OP² + PQ²
i.e., 1 + 2PQ = 7²
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
so, sin Q = 7/25 and cos Q = 24/25

```

Example-6 :-  In Δ ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (see Figure). Determine the lengths of the sides BC and AC.

Solution :-
```  To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB.
Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore
AB/BC = tan C
5/BC = tan 30°
5/BC = 1/√3
BC = 5√3 cm
To find the length of the side AC, we consider
sin 30° = AB/AC
1/2 = 5/AC
AC = 10cm
```

Example-7 :-  In Δ PQR, right-angled at Q (see Figure), PQ = 3 cm and PR = 6 cm. Determine ∠ QPR and ∠ PRQ.

Solution :-
```  Given PQ = 3 cm and PR = 6 cm
Therefore, PQ/PR = sin R
sin R = 3/6 = 1/2
so, ∠ PRQ = 30°
∠ QPR = 60°
```

Example-8 :-  If sin (A – B) = 1/2, cos (A + B) = 1/2, 0° < A + B ≤ 90°, A > B, find A and B.

Solution :-
```  Since, sin (A – B) = 1/2,
sin (A - B) = sin 30°
Therefore, A – B = 30° .....(1)
Also, since cos (A + B) = 1/2,
cos (A + B) = 60°
Therefore, A + B = 60° ......(2)
Solving (1) and (2),
A - B + A + B = 30° + 60°
2A = 90°
A = 90°/2
A = 45°
Put in (1) equation
A - B = 30°
45° - B = 30°
-B = 30° - 45°
-B = -15°
B = 15°
we get : A = 45° and B = 15°.
```

Example-9 :-  Evaluate : tan 65°/cot 25°

Solution :-
```  We know : cot A = tan (90° – A)
So, cot 25° = tan (90° – 25°) = tan 65°
i.e., tan 65°/cot 25° = tan 65°/tan 65° = 1
```

Example-10 :-  If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Solution :-
```  We are given that sin 3A = cos (A – 26°)..... (1)
Since sin 3A = cos (90° – 3A),
we can write (1) as cos (90° – 3A) = cos (A – 26°)
Since 90° – 3A and A – 26° are both acute angles,
Therefore, 90° – 3A = A – 26° which gives A = 29°
```

Example-11 :-  Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution :-
```  cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
```

Example-12 :-  Express the ratios cos A, tan A and sec A in terms of sin A.

Solution :-
```  Since cos² A + sin² A = 1,
Therefore, cos² A = 1 – sin² A,
i.e., cos A = √1 - sin² A
tan A = sin A/cos A = sin A/√1 - sin² A
sec A = 1/cos A = 1/√1 - sin² A
```

Example-13 :-  Prove that : sec A (1 – sin A)(sec A + tan A) = 1.

Solution :-
```
```

Example-14 :-  Prove that :

Solution :-
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Example-15 :-  Prove that : using the identity sec² θ = 1 + tan² θ.

Solution :-
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