ML Aggarwal Heights and Distances MCQs Solutions ICSE Class-10 Maths Ch-20. We Provide Step by Step Answer of MCQs Heights and Distances Questions for ICSE Class-10 APC Understanding Mathematics. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Heights and Distances MCQs Solutions ICSE Class-10 Maths Ch-20

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-20 | Heights and Distances |

Writer / Book | Understanding |

Topics | Solutions of MCQs |

Academic Session | 2024-2025 |

### Heights and Distances MCQs

**Choose the correct answer from the given four options (1 to 9):**

**Question 1. ****If a kite is flying at a height of 40 √3 metres from the level-ground, attached to a string inclined at 60° to the horizontal, then the length of the string is**

(a) 80 m

(b) 60 √3 m

(c) 80 √3 m

(d) 120 m

**Answer :**

Let K is kite

Height of KT = 40 √3 m

Angle of elevation of string at the ground = 60°

Let length of string AK = x m

Now sin 60° = KT/AK = (40√3)/x

√3/2 = (40√3)/x

x = (40√3 × 2)/√3 = 80 m

∴ Length of string = 80 m

**Question 2. ****If the angle of depression of an object from a 75 m high tower is 30°, then the distance of the object from the tower is**

(a) 25 √3 m

(b) 50√ 3 m

(c) 75 √3 m

(d) 150 m

**Answer :**

Height tower AB = 75 m

C is an object on the ground and angle of depression from A is 30°.

**Question 3. ****A ladder 14 m long rests against a wall. If the foot of the ladder is 7 m from the wall, then the angle of elevation is**

(a) 15°

(b) 30°

(c) 45°

(d) 60°

**Answer :**

Length of a ladder AB = 14 m

Foot of the ladder is 7 m from the wall θ is the angle of elevation

cos θ = BC/AB

= 7/14

= 1/2

= cos 60

= θ = 60**°**

**Question 4. ****If a pole 6 m high casts shadow 2 √3 m long on the ground, then the sun’s elevation is**

(a) 60°

(b) 45°

(c) 30°

(d) 90°

**Answer :**

Height of pole AB = 6 m

and its shadow BC = 2√3 m

3√3/√3 x √3

= 3√3/3

= √3 = tan 60**°**

θ = 60**°**

**Question 5. ****If the length of the shadow of a tower is √3 times that of its height, then the angle of elevation of the sun is**

(a) 15°

(b) 30°

(c) 45°

(d) 60°

**Answer :**

Let height of a tower AB = h m

Then its shadow BC = √3 hm

**Question 6. ****In ∆ABC, ∠A = 30° and ∠B = 90°. If AC = 8 cm, then its area is**

(a) 16 √3 cm²

(b) 16 m²

(c) 8 √3 cm²

(d) 6 √3 cm²

**Answer :**

In ∆ABC, ∠A = 30°, ∠B = 90°

AC = 8 cm

sin 30 = BC/AC

= 1/2 = BC/8

= BC = 8/2 = 4 cm

cos 30 = AB/AC

√3/2 = AB/8

= AB = 8√3/2 = 4√3 cm

near area triangle ABC = 1/2 AB x BC

= 1/2 x 4√3 x 4 cm²

= 8√3 cm²

— : End of ML Aggarwal Heights and Distances MCQs Solutions ICSE Class-10 Maths Ch-20 : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

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