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Very Short Answer Type Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Class 10 : Very Short Answer Type Questions- Introduction to Trigonometry Class 10 Notes | EduRev

The document Very Short Answer Type Questions- Introduction to Trigonometry Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Q1. If cotÎ¸ = 7/8 then what is the value of

Q2. If tan A   then find sin A.

Sol. In a right Î” ABC (âˆ B = 90Â°), Hypotenuse = AC, Base = AB and Perpendicular = BC.
Since,

Using Pythagoras theorem, we have:

Q3. Evaluate cos 60Â°Â· sin 30Â° + sin 60Â°Â· cos 30Â°.

Sol. We have:

cos 60Â°Â· sin 30Â° + sin 60Â°Â· cos 30Â°

Q4. If tan A = cot B, prove that A + B = 90Â°.

Sol. âˆµ tan A = cot B
âˆ´ tan A = tan (90Â° âˆ’ B)
â‡’ A = 90Â° âˆ’ B
â‡’ A + B = 90Â°.

Q5. Express sin 67Â° + cos 75Â° in terms of ratios of angles between 0Â° and 45Â°.
Sol.
âˆµ 67Â° = 90Â° âˆ’ 23Â° and 75Â° = 90Â° âˆ’ 15Â°

âˆ´ sin 67Â° + cos 75Â° = sin (90Â° âˆ’ 23Â°) + cos (90Â° âˆ’ 15Â°)  = cos 23Â° + sin 15Â°.

Q6. In the given figure, AC is the length of a ladder. Find it.

Sol. Let AC =x = [Length of ladder]
âˆ´ In right Î” ABC,

Thus, the length of the ladder is 2âˆš3 m

Q7. If sin Î¸ = 12/13 , find the value of:

Q8. In the given figure, find BC.

Sol. In Î” ABC,

Q9. In Î” ABC, if AD âŠ¥ BC and BD = 10 cm; âˆ  B = 60Â° and âˆ C = 30Â°, then find CD.

Sol. In right Î” ABD, we have

Q10. In the given figure,   ind AC, if AB = 12 cm.

Q11. In the given figure, Î” ABC is a right triangle. Find the value of 2 sinÎ¸ âˆ’ cosÎ¸.
Sol.
We have right Î” ABC,

Q12. In the figure, find sinA.

Sol. In right Î” ABC,

Q13. What is the value of sinÎ¸. cos(90Â° âˆ’ Î¸) + cosÎ¸ . sin(90Â° âˆ’ Î¸)?

Sol. sinÎ¸ Â· cos(90Â° âˆ’ Î¸) + cosÎ¸ Â· sin(90Â° âˆ’ Î¸) = sinÎ¸ Â· sinÎ¸ + cosÎ¸ Â· cosÎ¸
[âˆµ cos(90Â° âˆ’ Î¸) = sinÎ¸ , sin(90Â° âˆ’ Î¸) = cos Î¸]
= sin2 Î¸ + cos2 Î¸
= 1

Q14. Show that: tan 10Â° tan 15Â° tan 75Â° tan 80Â° = 1

Sol. We have:

L.H.S. = tan 10Â° tan 15Â° tan 75Â° tan 80Â° =
tan (90Â° âˆ’ 80Â°) tan 15Â° tan (90Â° âˆ’ 15Â°) tan 80Â°
= cot 80Â° tan 15 cot 15Â° tan 80Â° = (cot 80Â° Ã— tan 80Â°) Ã— (tan 15Â° Ã— cot 15Â°)
= 1Ã— 1
= 1 = R.H.S.

Q15. Find the value of:

Sol. We have:

Q16. Write the value of:

Q17. Write the value of:

Q18. If sec2 Î¸ (1 + sin Î¸) (1 âˆ’ sinÎ¸) = k, find the value of k.

Q19. If sin   then find the value of (2 cot2 Î¸ + 2).

Sol. 2 cot2 Î¸ + 2 = 2 (cot2 Î¸ + 1) = 2 (cosec2 Î¸)

Q20. If cos A = 3/5, find 9 cot2 A âˆ’ 1.

Q21. If sin 3Î¸ = cos (Î¸ âˆ’ 6)Â° and 3Î¸ and (Î¸ âˆ’ 6)Â° are acute angles, find the value of Î¸.

Sol. We have:
sin3Î¸ = cos(Î¸ âˆ’ 6)Â° = sin[90Â°âˆ’(Î¸ âˆ’ 6)Â°]

âˆµ [sin (90Â° âˆ’ Î¸) = cos Î¸]
â‡’ 3Î¸ = 90Â° âˆ’ (Î¸ âˆ’ 6)Â°
â‡’ 3Î¸ = 90 âˆ’ Î¸ + 6
â‡’ 3Î¸ + Î¸ = 96

Q22. If tan Î¸ = cot (30Â° + Î¸ ), find the value of Î¸ .

Sol. We have:
tan Î¸ = cot (30Â° + Î¸)
= tan [90Â° âˆ’ (30Â° + Î¸)]
= tan [90Â° âˆ’ 30Â° âˆ’ Î¸]
= tan (60Â° âˆ’ Î¸)
â‡’ Î¸ = 60Â° âˆ’ Î¸
â‡’ Î¸ + Î¸ = 60Â°

Q23. If sinÎ¸ = cosÎ¸ , find the value of Î¸ .

Sol. We have:
sinÎ¸ = cosÎ¸
Dividing both sides by cosÎ¸, we get

â‡’ tan Î¸ = 1 ...(1)
From , the table, we have:
tan 45Â° = 1 ...(2)
From (1) and (2), we have:
Î¸ = 45Â°.

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