Question

Two man who are 475 m apart, find the angles of elevation of a kite between them are 28 degree and 42 degree, respectively. what is the altitude of the kite, which is the same plane as them (math problem)

Answer 1

try draw it

Answer 2

can you also give a solution thankyou so much

Answer 3

thankyou so much can you also give a solution? that would be really help

Answer 4

the altitude of the kite let be h
on the laft part let be the man 1 and on the right part man 2
the distance between 1 man and the kite let be x and the distance between 2 man let be y
using these you can write

tan 28° = h/(475-y)

tan 42° = h/(475-x)

any idea now ?

Answer 5

or use @florisalreadytaken's idea with the law of sin

Answer 6

thankyou appreciated

Answer 7

so, law of sines tells us that:
\[\frac{\sin \alpha}{a}=\frac{\sin \gamma}{c} \]
look at the image i attached below, for the namng of the things.

Answer 8

\[\frac{\sin 42^o }{a}=\frac{\sin 110^o }{475} \Rightarrow a \sin \left(110^o \right)=475 \sin \left(42^o \right) \Rightarrow a=\frac{475 \sin \left(42^o \right) }{\sin \left(110^o\right)} \]
\[ a=?? \]

Answer 9

181?

Answer 10

no.

\[ a \approx 338.24 \]
the distance from the 1st guy to the kite is 338.24 m.

Answer 11

thankyou

Answer 12

by using SOH CAH TOA, we get:
\[ \sin(28^o) = \frac{opposite}{hypotenuse} \]
\[ \sin(28^o) = \frac{h}{a} \Rightarrow \sin(28^o) = \frac{h}{338.24} \]
\[ h=338.24 \sin(28^o) \]
\[ \Large h=?? \]

Answer 13

159

Answer 14

no...

\[ \Large h=158.63 \]

and that's tour answer

Answer 15

so thats the final ans?

Answer 16

positive

Answer 17

okay thankyou so much