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**Q.1. In Î” ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:****(i) sin A, cos A****(ii) sin C, cos C****Sol. **In right angled Î” ABC,

AC^{2} = AB^{2} + BC^{2}

[Pythagoras Theorem]

AC^{2} = (24)^{2} + (7)^{2}

= 576 + 49 = 625

â‡’ AC = 25 cm**Q.2. In the figure, find tan P âˆ’ cot R.****Sol.** In right angled Î”PQR,

PR^{2} = PQ^{2} + QR^{2}

[Pythagoras Theorem]

â‡’ (13)^{2} = (12)^{2} + QR^{2}

â‡’ 169 - 144 = OR^{2}

â‡’ 25 = OR^{2} â‡’ OR = 5 cm

So, **Q.3. If sin A = 3/4, calculate cos A and tan A.****Sol.**

Let BC = 3k

and AC = 4k

By Pythagoras theorem,**Q.4. Given 15 cot A = 8, find sin A and sec A.****Sol.** 15 cot A = 8

Let AB = Sk and BC = 15 k

In right angled Î”ABC,

AC^{2} = AB^{2} + BC^{2}

[Pythagoras theorem]

= (8k)^{2} + (15k)^{2}

= 64k^{2} + 225k^{2} = 289k^{2}

So,

and **Q.5. Given sec Î¸ = 13/12, calculate all other trigonometric ratios.****Sol.**

Let AC = 13k and AB = 12k

In right angled Î” ABC,**Q.6. If âˆ A and âˆ B are acute angles such that cos A = cos B, then show that âˆ A = âˆ B.****Sol. **Since **âˆ **A and **âˆ **B are acute angles.

Then, **âˆ **C = 90Â°

[Angles opposite to equal sides are equal]**Q.7. If cot Î¸ = 7/8, evaluate:****Sol.** cot Î¸ = 7/8

In right angled Î”ABC, AB/BC = 7/8

Let, AB = 7k and BC = 8k

In right angled Î”ABC, AC^{2} = AB^{2} + BC^{2}

[Pythagoras theorem]

**Q.8. If 3 cot A = 4, check whether = cos ^{2} A âˆ’ sin^{2} A or not.**

Let AC = 4k and BC = 3k

In right angled âˆ†ACB,

AB

= (4k)

â‡’ AB = 5k

In right angled Î”ABC,

AC

(i) sin A cos C + cos A sin

(ii) cos A cos C - sin A sin

In right angled Î”PQR

PR

â‡’ PQ

â‡’ (5)

â‡’ 25 = 25 (PR - QR)

â‡’ 25/25 = PR - OR

â‡’ PR - QR = 1 .... (i)

and PR + QR = 25 ... (ii)

On adding equation (i) and (ii), we get

From equation (i),

PR - QR = 1 â‡’ QR = 13 - 1 = 12 cm

(ii) sec A is always

(iii) cos A is the abbreviation for cosine A. (False)

(iv) cot without âˆ A is meaningless. (False)

(v) sin Î¸ can never be greater than 1.

âˆµ sin Î¸ = P/H, hypotenuse is always greater than other two sides.

Hence, the given statement is false.

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