Question

the the length of a string between a kite and a point on the ground is 93.5m. if the string makes an angle theta with the level ground such that tan theta=15/8, how high is the kite? please explain the s teps...​

Answer 1

AnsWer :

82.5 m.

GiveN :

• The length of string between a kite and a point on ground is 93.5 m.
• The value of tanθ = 15 / 8

SolutioN :

Let,

• The height of kite be 'H' meters = CA.
• The length of a string between a kite and a point on ground is 93.5 m = BC meters.

Now, We know that.

$\tt \bullet \: \: \: \: \: Sin\,\theta = \dfrac{Perpendicular}{Hypotenuse}$

$\tt \bullet \: \: \: \: \: tan\,\theta = \dfrac{Perpendicular}{Base} = \dfrac{15}{8}$

$\rule{200}3$

We need to Find the value of Sinθ

• So, Let's Apply [ Pythagoras theorem ]

• Mathematics :

$\tt \dagger \: \: \: \: \: H^2 = P^2 +B^2$

• Statement :

• One sides square of a triangle is equal to the sum of squares of other two sides.

where as,

• H = ?
• P = 15.
• B = 8.

$\tt : \implies \: H^2 = P^2 +B^2$

$\tt : \implies \: H^2 = (15)^2 +(8)^2$

$\tt : \implies \: H^2 =225+64.$

$\tt : \implies \: H^2 =289.$

$\tt : \implies \: H = \sqrt{289} .$

$\tt : \implies \: H = \sqrt{17 \times 17} .$

$\tt : \implies \: H =17.$

Now,

Other Method :

• We know that, 1 + tanθ = Secθ.

$\tt \dagger \: \: \: \: \: 1 +tan ^2\,\theta = Sec^2\,\theta.$

$\tt : \implies 1 + \Bigg\lgroup \dfrac{15}{8} \Bigg\rgroup^2 = Sec^2\,\theta.$

$\tt : \implies 1 + \Bigg\lgroup \dfrac{225}{64}\Bigg\rgroup = Sec^2\,\theta.$

$\tt : \implies \Bigg\lgroup \dfrac{64 + 225}{64} \Bigg\rgroup= Sec^2\,\theta.$

$\tt : \implies \Bigg\lgroup \dfrac{289}{64}\Bigg\rgroup = Sec^2\,\theta.$

$\tt : \implies Sec^2\,\theta = \dfrac{289}{64}$

$\tt : \implies Sec\,\theta = \sqrt{ \dfrac{289}{64} }$

$\tt : \implies Sec\,\theta = \dfrac{17}{8}$

$\rule{200}3$

Compare :

$\tt \bullet \: \: \: \: \: Sec\,\theta = \dfrac{hypotenuse}{Base} = \dfrac{17}{8}$

$\tt \bullet \: \: \: \: \: tan\,\theta = \dfrac{Perpendicular}{Base} = \dfrac{15}{8}$

Now,

• We have all value :
• H = 17.
• B = 8.
• P = 15.

$\rule{200}3$

★ Diagram :

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(15,-15){\line(0,0){25}}\put(-3.8,-15){\line(3,4){18.8}}\put(-3.8,-15){\line(5,0){18.8}}\qbezier(-2,-12.5)(0,-12.7)(0,-15)\put(12,-12){\line(1,0){3}}\put(12,-15){\line(0,1){3}}\put(1,-13){ \tt{\theta} }\put(14,-18){ \tt{A} }\put(14,11){ \tt{C} }\put(-6,-18){ \tt{B} }\end{picture}$

Let move to Question.

$\tt : \implies Sin\,\theta = \dfrac{Perpendicular}{Hypotenuse}$

• But, According to Question, we are let Perpendicular be H i.e height of kite from point on ground.
• similarly, Hypotenuse be length between a kite and ground is 93.5 m = BC m.

$\tt : \implies Sin\,\theta = \dfrac{CA}{BC}$

$\tt : \implies Sin\,\theta = \dfrac{H}{93.5}$

$\tt : \implies \dfrac{15}{17} = \dfrac{H}{93.5}$

$\tt : \implies H=\dfrac{15 \times 93.5}{17}$

$\tt : \implies H=82.5 \: m.$

Therefore, the Height of kite from point on ground be 82.5 m.

Answer 2

Explanation:

AnsWer :

82.5 m.

GiveN :

The length of string between a kite and a point on ground is 93.5 m.

The value of tanθ = 15 / 8

SolutioN :

Let,

The height of kite be 'H' meters = CA.

The length of a string between a kite and a point on ground is 93.5 m = BC meters.

Now, We know that.