Given:
A plane is 142 km north and 166 km east of an airport as shown in the given figure.
To find:
The value of x to the nearest tenth of a degree.
Solution:
In a right angle triangle,
[tex]\tan \theta=\dfrac{Perpendicular}{Base}[/tex]
It is also written as
[tex]\tan \theta=\dfrac{Opposite}{Adjacent}[/tex]
For the given triangle,
[tex]\tan x=\dfrac{142}{166}[/tex]
[tex]\tan x=\dfrac{71}{83}[/tex]
[tex]x=\tan^{-1} \dfrac{71}{83}[/tex]
[tex]x\approx 40.5[/tex]
Therefore, the value of x is 40.5 degrees.
