Question

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A boy standing at a point O finds his kite flying at point P with distance OP = 25 m. It is a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground. ( use trigonometric ratios )

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AN⛎SHA ☺❤✌❄​

Answer 1

Hey Miss

Here's The Answer

_______________________________

Let Angle POM be ∅.

Given :

PM = 5 m

OP = 25 m

PQ = 10 m

To Find : QN ( h )

By figure it's clear, Angle POM = Angle QON

From ∆ POM

=> sin ∅ = PM / OP

=> sin ∅ = 5 / 25

=> sin ∅ = 1 / 5 _______ ( i )

From ∆ QON

=> sin ∅ = QN / OQ

=> sin ∅ = h / ( OP + PQ )

=> sin ∅ = h / ( 25 + 10 )

=> sin ∅ = h / 35 _____ ( ii )

Now equating ( i ) and ( ii )

=> sin ∅ = 1 / 5 = h / 35

=> 1/5 = h/35

=> h = 35/5

=> h = 7 m

Hence QN = 7 meters

Hope It Helps ^^

Answer 2

$\bf{\underline{\underline{Question:-}}}$

A boy standing at a point O finds his kite flying at point P with distance OP = 25 m. It is a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground. ( use trigonometric ratios )

$\bf{\underline{\underline{Find:-}}}$

• Height of QN of the kite from the ground ?

$\bf{\underline{\underline{Solution:-}}}$

◆ In Δ OMP and Δ ONQ

◆ ∠O is common.

◆ ∠OMP = ∠ON Q ( each 90° )

◆ ΔOMP ~ Δ ONQ = [AA Similarity]

★ As when triangles are similar ,

• their Sides are proportional.

→ $\sf \dfrac{OP}{OQ} = \dfrac{PM}{QN}$

→$\sf \dfrac{25}{35}= \dfrac{5}{QN}$

→$\sf \dfrac{35×5}{25}=QN$

→$\sf \dfrac{175}{25}=QN$

→$\sf 7=QN$

$\bf{\underline{\underline{Hence:-}}}$

• Height of QN of the kite from the ground is 7m