Courses

# Short Answer Type Questions(Part- 2)- Introduction to Trigonometry Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions(Part- 2)- Introduction to Trigonometry Class 10 Notes | EduRev

The document Short Answer Type Questions(Part- 2)- Introduction to Trigonometry Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

Q16. For an acute angle Î¸ , show that: (sin Î¸ âˆ’ cosec Î¸) (cos Î¸ âˆ’ sec Î¸)

Q20. Without using trigonometric tables evaluate:

Q21. If tan (A + B) = âˆš3 and tan (A âˆ’ B) = 1, 0Â° < A + B < 90Â°; A > B, then find A and B.

Sol. We have: tan (A + B) = âˆš3 (Given)
tan 60Â° = âˆš3 (From the table)
â‡’ A + B = 60Â° ...(1)
Also,tan (A âˆ’ B)= 1 [Given]
and cosec 60Â° = 2 and cos 90Â° = 0
â‡’ A âˆ’ B = 45 ...(2)
2A = 60Â° + 45Â° = 105Â°

â‡’

From (2), 52.5Â° âˆ’ B = 45Â°
â‡’ B = 52.5Â° âˆ’ 45Â° = 7.5Â°
Thus, A = 52.5Â° and B = 7.5Â°.

Q22. If tan (2A) = cot (A âˆ’ 21Â°), where 2A is an acute angle, then find the value of A.

Sol. We have: tan (2A) = cot (A âˆ’ 21Â°)
âˆµ cot (90Â°âˆ’ Î¸) = tan Î¸
âˆ´ cot (90Â°âˆ’ 2A) = tan 2A
â‡’ cot (90Â°âˆ’ 2A) = cot (A âˆ’ 21)Â°
â‡’ 90 âˆ’ 2A = A âˆ’ 21Â°
â‡’ âˆ’ 2A âˆ’ A = âˆ’ 21Â°âˆ’ 90Â°
â‡’ âˆ’ 3A = âˆ’ 111Â°

â‡’

Q23. If sin 3A = cos (A âˆ’ 10Â°), then find the value of A, where 3A is an acute angle.

Sol. We have:
sin 3A = cos (A âˆ’ 10Â°)
âˆµ cos (90Â° âˆ’ Î¸) = sin Î¸
âˆ´ cos (90Â° âˆ’ 3A) = sin 3A
â‡’ 90Â° âˆ’ 3A = A âˆ’ 10Â°
â‡’ âˆ’3A âˆ’ A = âˆ’10Â° âˆ’ 90Â°
â‡’ âˆ’4A = âˆ’100Â°

â‡’

Q24. If sec 2A = cosec (A âˆ’ 27Â°), then find the value of A, where 2A is an acute angle.

Sol. We have:

sec 2A = cosec (A âˆ’ 27Â°) ...(1)
âˆµ sec Î¸ = cosec (90Â° âˆ’ Î¸)
âˆ´ sec 2A = cosec (90Â° âˆ’ 2A) ...(2)

From (1) and (2), we get
A âˆ’ 27Â° = 90Â° âˆ’ 2A
â‡’ A + 2A = 90 + 27Â° = 117Â°
â‡’ 3A = 117Â°
â‡’

Q25. Simplify:

+ sin Î¸ cos Î¸

Sol. We have:

= 1 = R.H.S.

Q28. Without using trigonometrical tables, evaluate:

Sol.

[âˆµ sin (90Â° âˆ’ Î¸) = cos Î¸, cos (90Â° âˆ’ Î¸) = sin Î¸, cosec (90Â° âˆ’ Î¸) = sec Î¸, and tan (90Â° âˆ’ Î¸) = cot Î¸]

Q29. Using Geometry, find the value of sin 60Â°.

Sol. Let us consider an equilateral Î”ABC and draw AD âŠ¥ BC.
Since, each angle of an equilateral triangle = 60Â°
âˆ´âˆ A = âˆ B = âˆ C = 60Â°

Let AB = BC = AC = 2a

In Î”ABD and Î”ACD, we have:
AB = AC   [Given]
â‡’ Î”ABD â‰… Î”ACD
â‡’ BD = CD

Now, using Pythagoras theorem, in right Î”ABD,

= (2a)2 âˆ’ a2
= 4a2 âˆ’ a2
= 3a2

â‡’

âˆ´

Thus,

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;