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Compulsory Mathematics /275

एकजना केटा चड्गा उडाइ रहेको छ। हावामा धागोको लम्बाइ 300 m हुँदा धागोले क्षितिजसंग [tex]$45^{\circ}$[/tex] को कोण बनाएको छ।

A boy is flying a kite. The string is 300 m long and formed an angle of [tex]$45^{\circ}$[/tex] with the horizon.

(a) [tex]$\cos 45^{\circ}$[/tex] को मान लेखुनुहोस् ।
(Write the value of [tex]$\cos 45^{\circ}$[/tex].)

(b) AB को नाप पत्ता लगाउनुहोस् ।
(Find the measure of AB.)

(c) यदि केटाको उचाइ 2 m भए जमिनबाट सो चड्गा कति माथि होला ?
If the height of the boy is 2 m, then what is the height of the kite from the ground level?

(d) हावाको कारणले उन्नतांश कोण [tex]$45^{\circ}$[/tex] बाट [tex]$30^{\circ}$[/tex] मा परिवर्तन भयो। अब चड्गाको उचाइ जमिनबाट कति मुनि होला ?
Due to airflow, the angle of elevation changes from [tex]$45^{\circ}$[/tex] to [tex]$30^{\circ}$[/tex]. What is the kite's height from the ground level now?

Answers:
(b) 212.13 m
(c) 214.13 m
(d) 152 m



Answer :

GinnyAnswer
Certainly! Let's solve the problem step-by-step.

### Given:
- Length of string (L) = 300 meters
- Initial angle with the horizon = [tex]\( 45^{\circ} \)[/tex]
- Height of the boy (h) = 2 meters
- Angle after change due to airflow = [tex]\( 30^{\circ} \)[/tex]

### (a) Value of [tex]\( \cos 45^{\circ} \)[/tex]:

The value of [tex]\( \cos 45^{\circ} \)[/tex] is well-known and is:
[tex]\[ \cos 45^{\circ} = 0.7071 \][/tex]

### (b) Finding the measure of AB:

AB is the horizontal distance from the boy to the point directly below the kite. Using the trigonometric relationship:
[tex]\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \][/tex]
where [tex]\(\theta\)[/tex] is the angle with the horizontal and the hypotenuse is the length of the string (L).

For [tex]\(\theta = 45^{\circ}\)[/tex]:
[tex]\[ \cos 45^{\circ} = \frac{AB}{L} \][/tex]
[tex]\[ AB = L \cos 45^{\circ} \][/tex]
Substituting the values:
[tex]\[ AB = 300 \times 0.7071 = 212.13 \text{ meters} \][/tex]

### (c) Height of the kite from the ground level when the angle is [tex]\(45^{\circ}\)[/tex]:

We calculate the vertical height of the kite using:
[tex]\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \][/tex]
For [tex]\(\theta = 45^{\circ}\)[/tex]:
[tex]\[ \sin 45^{\circ} = \frac{\text{Height above the boy}}{L} \][/tex]
[tex]\[ \text{Height above the boy} = L \sin 45^{\circ} \][/tex]

Since [tex]\(\cos 45^{\circ} = \sin 45^{\circ}\)[/tex]:
[tex]\[ \sin 45^{\circ} = 0.7071 \][/tex]

Thus:
[tex]\[ \text{Height above the boy} = 300 \times 0.7071 = 212.13 \text{ meters} \][/tex]

Total height from the ground:
[tex]\[ \text{Height from the ground} = \text{Height above the boy} + \text{boy's height} \][/tex]
[tex]\[ \text{Height from the ground} = 212.13 + 2 = 214.13 \text{ meters} \][/tex]

### (d) Height of the kite from the ground level when the angle is [tex]\(30^{\circ}\)[/tex]:

For [tex]\(\theta = 30^{\circ}\)[/tex]:
[tex]\[ \sin 30^{\circ} = \frac{\text{Height above the boy}}{L} \][/tex]
[tex]\[ \sin 30^{\circ} = 0.5 \][/tex]

Thus:
[tex]\[ \text{Height above the boy} = 300 \times 0.5 = 150 \text{ meters} \][/tex]

Total height from the ground:
[tex]\[ \text{Height from the ground} = \text{Height above the boy} + \text{boy's height} \][/tex]
[tex]\[ \text{Height from the ground} = 150 + 2 = 152 \text{ meters} \][/tex]

### Summary:
(a) The value of [tex]\( \cos 45^{\circ} \)[/tex] is 0.7071.

(b) The horizontal distance AB is 212.13 meters.

(c) The height of the kite from the ground level when the angle is [tex]\( 45^{\circ} \)[/tex] is 214.13 meters.

(d) The height of the kite from the ground level when the angle changes to [tex]\( 30^{\circ} \)[/tex] is 152 meters.

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