Trigonometry Questions for Competitive Exams Part 12th

Following the sequence of last post on basics of Trigonometry I am continuing the basics with their related questions in this post. Let’s prepare this to get full hold on the questions of trigonometry.

Q22. If Cosθ =5/13 and θ is an acute angle [0< θ < π/2], then find the value of

            [{Cosθ + 5Cotθ}/{Cosecθ – Cosθ}].

Sol.  Given :-  Cosθ =5/13

                                                                                                                                                                                                                                                                                                                                                                                                

In triangle ABC with 90⁰ at B, Cosθ = BC/CA = 5/13

This means Base (BC) = 5 units and Hypotenuse (AC) = 13 units

As we have seen and discussed in the last post of Trigonometry, the important triplet numbers which always form a Right angled Triangle. Let’s Revise it

3,4 and 5 (3 and 4 will be measures of base or perpendicular and 5 will be hypotenuse.)

6,8 and 10 (6 and 8 will be measures of base or perpendicular and 10 will be hypotenuse.)

12,5 and 13. (12 and 5 will be measures of base or perpendicular and 13 will be hypotenuse.)

15,8 and 17. (15 and 8 will be measures of base or perpendicular and 17 will be hypotenuse.)

24, 7 and 25. (24 and 7 will be measures of base or perpendicular and 25 will be hypotenuse.)

With the help of coloured triplet we get

Perpendicular (AB) = 12 units.

Now, Cosecθ = AC/AB = 13/12

And Cotθ = BC/AB = 5/12

Now according to the question

[{Cosθ + 5Cotθ}/{Cosecθ – Cosθ}].

Now on putting the values of Cosθ, Cosecθ and Cotθ from above we get;

= [{(5/13) + 5(5/12)}/{(13/12) – (5/13)}]

= [{(5/13) + (25/12)}/{(13/12) – (5/13)}]

On taking LCM and Solving we get

= [{(60 + 325)/(13 × 12)}/(169 - 60)/(13 × 12)}]

On solving we get,

= 385/109.

 Hence the required answer is 385/109.

Q23. If Cosθ = 5/13, then the value of (tan²θ + Sin²θ) = ?  

Sol. Given :-  Cosθ =5/13

In triangle ABC with 90⁰ at B, Cosθ = BC/CA = 5/13

This means Base (BC) = 5 units and Hypotenuse (AC) = 13 units

As we have seen and discussed in the last post of Trigonometry, the important triplet numbers which always form a Right angled Triangle. Let’s Revise it

3,4 and 5 (3 and 4 will be measures of base or perpendicular and 5 will be hypotenuse.)

6,8 and 10 (6 and 8 will be measures of base or perpendicular and 10 will be hypotenuse.)

12,5 and 13. (12 and 5 will be measures of base or perpendicular and 13 will be hypotenuse.)

15,8 and 17. (15 and 8 will be measures of base or perpendicular and 17 will be hypotenuse.)

24, 7 and 25. (24 and 7 will be measures of base or perpendicular and 25 will be hypotenuse.)

With the help of coloured triplet we get

Perpendicular (AB) = 12 units.

Now, tanθ = AB/BC = 12/5

And Sinθ = AB/AC = 12/13

Now according to the question

tan²θ + Sin²θ

Now on putting the values of tanθ and Sinθ from above we get;

= (12/5)² + (12/13)²

= (144/25) + (144/169)

= 144 [(1/25) + (1/169)]

= 144 [(169 + 25)/(25 × 169)]

= (144 × 194)/(25 × 169)

= 27936/4225.

Hence the required answer is 27936/4225.

Q24. Find the value of Sin15⁰ .

Sol. Given :-  Sin15⁰

= Sin(45⁰ - 30⁰)

We know that Sin(A - B) = SinA . CosB - CosA . SinB, on applying this formula;

= Sin45⁰ . Cos30⁰ – Cos45⁰ . Sin30⁰

= (1/√2) . (√3/2) – (1/√2) . (1/2)

= (√3/2√2) – (1/2√2)

= (√3 - 1)/2√2

 Hence the required answer Sin15⁰ = (√3 - 1)/2√2.

 (Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)                                                                        

Hope you have enjoyed the base concept of this chapter till now. More Conceptual enjoyment will be continued for you soon, just keep visiting this blog regularly and writing me your related queries through comment or forum.

Best of Luck for Exams. To be continued soon........................