Answer:
∴ [tex]\tan \theta = -\frac{3}{4}[/tex]
Step-by-step explanation:
Let the triangle name as Δ ABO a right triangle at ∠ B =90°
Such that,
OA = radius
AB = y coordinate = 12
BO = x coordinate = 16 (positive because distance cannot be in negative)
To Find:
[tex]\tan \theta=?[/tex]
Solution:
In right triangle Δ ABO ,By Pythagoras theorem we get the radius,
[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex]
[tex]r^{2} = 16^{2}+12^{2}\\\\r^{2} =400\\\therefore r = 20\ units[/tex]
∴ OA = 20
OB = 16
Also tan (180 -θ) = - tan (θ)
Now In right triangle Δ ABO
m∠ AOB = 180 -θ
∴ [tex]\tan (\angle AOB) = \frac{\textrm{side opposite to angle AOB}}{\textrm{side adjacent to angle AOB}}\\\\\tan (180 -\theta) = \frac{AB}{OB}\\\\-\tan \theta =\frac{12}{16} \\\\\tan \theta = -\frac{3}{4}[/tex]
∴ [tex]\tan \theta = -\frac{3}{4}[/tex]
