Verified answer

Given :-

• A kite flies in the sky with a thread of 68m and makes an angle θ
• Tan θ = 15/8

To Find :-

• What is the height of the kite to above the ground ?

Formulas Used :-

[tex] \displaystyle \: \bf \: \implies \: Tan \: θ \: = \: \frac{Opposite \: side}{Adjacent \: side} [/tex]

[tex] \: \bf \implies \: Hypotenuse^{2} \: = \: Base^{2} + Height^{2} [/tex]

Solution :-

Given,

[tex]⇝[/tex]A kite flies in the sky with a thread of 68m and makes an angle θ

[tex]⇝[/tex]Tan θ = 15/8

We know that,

[tex]\displaystyle \: \sf \: \implies \: Tan \: θ \: = \: \frac{Opposite \: side}{Adjacent \: side} [/tex]

[tex]\displaystyle \: \sf \: \implies \: Tan \: θ \: = \: \frac{15}{8} [/tex]

Let,

• Hypotenuse = 68m
• Height = 15x
• Base = 8x

By Pythagoras Theorem,

[tex] \: \sf \implies \: Hypotenuse^{2} \: = \: Base^{2} + Height^{2} [/tex]

[tex] \: \sf \implies \: (68)^{2} \: = \: (8x)^{2} + (15x)^{2} [/tex]

[tex] \: \sf \implies \: 4624\: = \: 64 x^{2} + 225x^{2} [/tex]

[tex] \: \sf \implies \: 4624\: = \: 289x^{2} [/tex]

[tex] \displaystyle \: \sf \implies \: \cancel\frac{4624}{289} \: = \: x^{2} [/tex]

[tex]\displaystyle \: \sf \implies \: \frac{16}{1} \: = \: x^{2} [/tex]

[tex]\displaystyle \: \sf \implies \: 16 \: = \: x^{2} [/tex]

[tex]\displaystyle \: \sf \implies \: 4 \: = \: x[/tex]

[tex]\displaystyle \: \bf \implies \: x \: = \: 4[/tex]

Hence,

∴ Height of kite from the ground = 15(4) = 60m

[tex] \: [/tex]