**Q1. A tower standing at a horizontal place makes a certain angle at a point 160 m away from its foot. When the point is moved towards the foot of the tower by 100, the angle subtended by the tower at this point becomes two times the previous angle. Accordingly, the height of tower will be –**

(a) 80 m

(b) 100 m

(c) 160 m

(d) 200 m

**Q2. If a² + b² + c² = 2(a – b – c) – 3, the value of 4a – 3b+ 5c will be –**

(a) 2

(b) 3

(c) 5

(d) 6

**Q3. In ∆ABC, AD⊥BC. E is a point on AD such that AE : ED = 5 : 1. Accordingly if ∠BAD = 30° and tan (∠ACB) = 6 tan (∠DBE), find ∠ACB = ?**

(a) 30°

(b) 45°

(c) 60°

(d) 15°

**Q4. An equilateral ∆BEC is inscribed in a square ABCD. CE and BD intersect each other at point O, then ∠BOC will be equal to –**

(a) 60°

(b) 75°

(c) 90°

(d) 120°

**Q5. A right pyramid is based on an equilateral triangular base whose side is 10√3 cm long. If the total surface area of the pyramid is 270√3 cm², its height will be –**

(a) 12√3 cm

(b) 10 cm

(c) 10√3 cm

(d) 12 cm

**Q6. A circular park whose radius is 20 m long is situated in a colony. Three boys namely Ankur, Syed and David are sitting at its circumference at equal distance and each has a toy telephone in his hand to talk. Find the length of the chord of the each telephone.**

(a) 40√3m

(b) 10m

(c) 20√3m

(d) 15√2m

**Q7. The common chord of two circles is 30 cm long. If the diameters of the circles are 50 cm and 34 cm long, find the distance between their centres.**

(a) 14 cm

(b) 28 cm

(c) 13 cm

(d) 30 cm

**Q8. The base of a pyramid is an equilateral triangular shape. Its volume is 48√3 cm² and height is 4 cm long. Find the side of the base.**

(a) 4 cm

(b) 6 cm

(c) 12 cm

(d) 3 cm

**Q9. Point O is the in-centre of ∆ABC and the circle inscribed in it touches the sides BC, CA and AB at points P, Q and R respectively. If ∠A= 60°, find the value of ∠QPR. **

(a) 90°

(b) 120°

(c) 100°

(d) 60°

(b) 0

(c) 1

(d) 2

**Q11. If a² + b² + c² = ab + bc + ca, find the value of (a+c)/b.**

(a) 3

(b) 4

(c) 1

(d) 2

**Q12. If 3sinθ + 5cosθ = 5, then the value of 5sinθ – 3cosθ.**

(a) ± 3

(b) ± 5

(c) 1

(d) ±2

**Q13. The base of a right prism is a right angled triangular shape, whose sides are 5 cm, 12 cm and 13 cm long. If the total surface area of the prism is 360 cm², find its height in cm.**

(a) 10

(b) 12

(c) 9

(d) 11

**Q14. If tanθ = (sinα-cosα)/(sinα+cosα,)the value of sinα + cosα is –**

(a) ± √2 sinθ

(b) ± √2 cosθ

(c) ±1/√2 sinθ

(d) ±1/√2 cosθ

(a) 0

(b) 1

(c) 3

(d) a + b + c

**Solutions**

**S2. Ans.(a)**

**Sol. ∵ a² + b² + c² = 2(a – b – c) – 3**

**⇒ a² – 2a + 1 + b² + 2b + 1 + c² + 2c + 1 = 0**

**⇒ (a – 1)² + (b + 1)² + (c + 1)² = 0**

**It is only possible when—**

**(a – 1)² = 0, (b + 1)² = 0, (c + 1)² = 0**

**∴ a = 1, b = –1 & c = –1**

**∴ 4a – 3b + 5c = 4 × 1 – 3(–1) + 5(–1)**

**= 4 + 3 – 5 = 2**

**S11. Ans.(d)**

**Sol. ∵ a² + b² + c² = ab + bc + ca**

**∴ 2a² + 2b² + 2c² = 2ab + 2bc + 2ca**

**or, a² – 2ab + b² + b² – 2bc + c² + c² – 2ac + a² = 0**

**or, (a – b)² + (b – c)² + (c – a)² = 0**

**⇒ (a – b)² = 0 ⇒ a = b**

**⇒ (b – c)² = 0 ⇒ b = c**

**∴ a = b = c**

**∴(a+c)/b=(b+b)/b=2**

**S12. Ans.(a)**

**Sol. Let 5 sin θ – 3 cos θ = x …(i)**

**from question,**

**3 sin θ + 5 cos θ = 5 …(ii)**

**By squaring and adding (i) & (ii)**

**25 + 9 = x² + 25**

**⇒ x = ±3**

**S13. Ans.(a)**

**Sol. Let height = h**

**∴ 360 = 2 × 1/2 × 12 × 5 + [5 × h + 12 × h + 13× h]**

**∴ 360 = 60 + 30h**

**∴ h = 10 cm**

**∴ height = 10 cm**

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