## Trigonometry & Height and Distances

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Question 1 |

What is the maximum value of 5 Sinθ + 12 cosθ?

16 | |

13 | |

10 | |

12 |

**Trigonometry & Height and Distances**

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Question 1 Explanation:

Maximum value = √(a

^{2}+ b^{2}) =13Question 2 |

What is minimum value of Sinθ + cosθ ?

-2 | |

1/2 | |

-1 | |

-√2 |

**Trigonometry & Height and Distances**

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Question 2 Explanation:

Minimum value = √(a

=-√2

^{2}- b^{2})=-√2

Question 3 |

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

2/√3 | |

1/2 | |

1/√2 | |

1/√3 |

**Trigonometry & Height and Distances**

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Question 3 Explanation:

tanA = cotB,

tanA*tan B = 1

So, A +B = 90

(x+y)+(x-y) = 90

Tan (2x/3) = tan 30

tanA*tan B = 1

So, A +B = 90

^{o}(x+y)+(x-y) = 90

^{o}, 2x = 90^{o}, x = 45^{o}Tan (2x/3) = tan 30

^{o}= 1/√3Question 4 |

Find the Value of tan60

^{o}+ tan120^{o}?1/2 | |

0 | |

2 | |

2/3 |

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Question 4 Explanation:

=tan(60) + tan(180-120)
=tan(60)+(-tan(60)) {because in second quadrant tan is -ve}
=0

Question 5 |

The least value of 2sin

^{2}θ + 3cos^{2}θ1/3 | |

4/3 | |

2 | |

3/4 |

**Trigonometry & Height and Distances**

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Question 5 Explanation:

= 2sin

=2(sin

= 2 + cos

^{2}θ + 2cos^{2}θ + cos^{2}θ=2(sin

^{2}θ + cos^{2}θ) + cos^{2}θ ; (by putting sin^{2}θ + cos^{2}=1)= 2 + cos

^{2}θ ;(the minimum value of cos^{2}θ=0) = 2 + 0 = 2Question 6 |

The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree, is:

30 degrees | |

45 degrees | |

60 degrees | |

90 degrees |

**Trigonometry & Height and Distances**

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Question 6 Explanation:

Let AB be the height of the tree and AC be the length of the shadow. We need to calculate the angle ACB where BC is the hypotenuse.
Given,
AC:AB = √3
cot θ = √3
θ = 30 degrees.

Question 7 |

From a point P on a level ground, the angle of elevation of the top of a tower is 30 degrees. If the tower is 100 m high, the distance of point P from the foot of the tower is:

149 m | |

156 m | |

173 m | |

200 m |

**Trigonometry & Height and Distances**

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Question 7 Explanation:

Let QR be the tower. Then,
QR = 100 m and angle QPR = 30 degrees.
We know, cot 30° = √3 = PQ/QR.
Therefore, PQ = 100*√3 = 173 m.

Question 8 |

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man walks some distance towards the tower to watch its top and the angle of elevation becomes 60°. What is the distance between the base of the tower and the point P?

4(3)^0.5 units | |

8 units | |

12 units | |

Data inadequate |

**Trigonometry & Height and Distances**

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Question 8 Explanation:

Let MN be the tower and man be standing at P (30° = angle MPN) and Q (60° = angle MQN). We are only given two angles and no sides of the triangles. Therefore, the data is inadequate.

Question 9 |

A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 meters from the tower. After 5 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?

32 kmph | |

36 kmph | |

40 kmph | |

44 kmph |

**Trigonometry & Height and Distances**

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Question 9 Explanation:

Let the tower be PQ and the boat be at positions R and S when making angles of 45° and 30° respectively.
Given, PR = 60 m.
Now, PQ/PR = tan 45° = 1. So, PQ = PR = 60 m.
Again, PQ/PS = tan 30° = 1/√3. So, PS = 60 * √3 m = 103.92 m.
Distance covered in 5 seconds = 103.92 - 60 = 43.92 m.
Speed in kmph = (43.92/5) * (18/5) = 32 kmph (approximately)

There are 9 questions to complete.