Question

Hector is flying a kite. He has let out 86 feet
of string and is holding it 4 feet off the
ground. If the string is at an angle of
elevation of 42°15’30”, how high is the kite?

Answer 1

Answer:

Hector's kite is 61.84 feet from the ground.

Step-by-step explanation:

The angle of elevation of the kite is 42°15’30” when converted to decimals, it is $42.258333^{0}$ ≅ $42.26^{0}$

Let the height of the kite to the horizontal of angle of elevation be represented as x. Applying the trigonometric function to the sketch of Hector's kite,

Sin θ = $\frac{opposite}{hypotenus}$

Sin $42.26^{0}$ = $\frac{x}{86}$

⇒ x = 86 x Sin $42.26^{0}$

      = 86 x 0.6725

     = 57.835

x ≅ 57.84 feet

The height of Hector's kite from the ground = x + 4

                                    = 57.84 + 4

                                   = 61.84 feet