Question

A kite if flying at a height of 75m from the level ground, attached to a string inclined at 60 to the horizontal, Find the length of the string to the nearest metre.
(only for mods/stars/best users)
(saby123, copythat, brainlycommando)
please answer this question.

Answer 1

Answer:

• 86.6m.

Step-by-step explanation:

Let OX be the horizontal line and A be the position of the kite with string OA.

Draw AB ⊥ OX.

Then, AB = 75 m and ∠BOA = 60°.

From right ΔOBA, we get:

AB/OA = sin 60° ⟹ 75/OA = √3/2

⟹ OA = (75 × 2/√3 × √3/√3)m

⟹ OA = (150 × 1/√3 × √3)m

⟹ OA = (50 × √3)m

⟹ OA = (50 × 1.732)m = 86.6 m.

• Hence, the length of the string to the nearest metre is 86.6 m.

Answer 2

Given : A kite is flying at a height of 75 metre from the level ground with an attached string inclined at 60° from the horizon.

To find : The length of the string

Solution :

The given problem is an application of trigonometry. We will solve this question by using the concept of trigonometric ratios involving with hypotenuse and height of a triangle.

Let's suppose a point A be the exact position of the kite and AC be the string of x metres. Construct a perpendicular from the kite to the horizon say it AB .

Since the height of the kite from the ground is given to be 75 m, length of AB will be 75 metres . Also it is given that kite is inclined at an angle of 60° implies that ∠ ACB = 60° .

Now this formed a right angled triangle where right angle at B. We have to find the length of AC. [ See the attachment ].

We are aware about the sin θ, trigonometric ratio involving with the hypotenuse and height of the triangle.

Consider ∆ ABC :-

→ sin θ = Perpendicular / Base

→ sin 60° = AB / AC

Now substitute value of sin 60° , AB and AC .

→ √3 / 2 = 75 m / x

→ x = (75 × 2 m) / √3

→ x = 150 m / √3

→ x = 50 √3 metres

Since the value of x is 50√3 m , required length of the string is 50 √3 m.

Basic trigonometric formulas :-

• sin A = Perpendicular / Hypotenuse
• cos A = Base / Hypotenuse
• tan A = Perpendicular / base
• cosec A = Hypotenuse / Perpendicular
• cos A = Hypotenuse / base
• cot A = Base / Hypotenuse

• sec² A - tan² A = 1
• 1 + cot² A = cosec²A
• sin² A + cos² A = 1

• sin A = 1 / cosec A
• cos A = 1 / sec A
• tan A = 1 / cot A
• cosec A = 1 / sin A
• sec A = 1 / cos A
• cot A = 1 / tan A
• tan A = sin A / cos A
• tan A = sec A / cosec A
• cot A = cos A / sin A
• cot A = cosec A / cos A

• sin ( 90° - A ) = cos A
• cos ( 90° - A ) = sin A
• tan ( 90° - A ) = cot A
• cot ( 90° - A ) = tan A
• cosec ( 90° - A ) = sec A
• sec ( 90° - A ) = cos A