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.

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**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) = **âˆš**3 (Given)

tan 60Â° = **âˆš**3 (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)

Adding (1) and (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]

âˆ ADB = âˆ ADC = 90Â° [Construction]

AD = AD [Construction]

â‡’ Î”ABD â‰… Î”ACD

â‡’ BD = CD

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

AD^{2} = AB^{2} âˆ’ BD^{2}

= (2a)^{2} âˆ’ a^{2}

= 4a^{2} âˆ’ a^{2}

= 3a^{2}

â‡’

âˆ´

Thus,

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