GIVEN :-
Coordinates of points are :-
(-5 , 12)(2 , 8)(3 , -6)
TO FIND :-
All the trigonometric values of given points
FACTS TO KNOW BEFORE SOLVING :-
It's important to know that :-
In 1st quadrant (0° to 90°) , all the trigonometric values are positive .In 2nd quadrant (90° to 180°) , except sin & cosec , rest all trigonometric values are negative.In 3rd quadrant (180° to 270°) , except tan & cot , rest all trigonometric values are negative.In 4th quadrant (270° to 360°) , except cos & sec , rest all all trigonometric values are negative.
SOLUTION :-
Q1)
Plot (-5,12) on the cartesian plane and name it 'A'Drop a perpendicular to x-axis from 'A' and name the point 'B' where the perpendicular meets x-axis.Join the point A with the origin 'O'.
You'll notice a right-angled ΔABO formed (∠B = 90°) in the 2nd quadrant of the plane whose :-
length of perpendicular of triangle (AB) = 12 unitslength of base of triangle (OB) = 5 unitslength of hypotenuse (OA) = 13 units
Let the angle between OA & positive x-axis be θ.
⇒ ∠AOB = 180 - θ
So ,
![\sin (AOB) = \sin(180 - \theta) = \sin \theta = \frac{12}{13}](/tpl/images/1323/3619/6b51f.png)
![\cos(AOB) = \cos (180 - \theta) = -\cos \theta = -\frac{5}{13}](/tpl/images/1323/3619/7d635.png)
![\tan(AOB) = \tan(180 - \theta) = -tan \theta = -\frac{12}{5}](/tpl/images/1323/3619/ef0fa.png)
![\csc(AOB) = \csc(180 - \theta) = \csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}](/tpl/images/1323/3619/93b8e.png)
![\sec(AOB) = \sec(180 - \theta) = -\sec \theta = -\frac{1}{\cos \theta} = -\frac{13}{5}](/tpl/images/1323/3619/8c92f.png)
![\cot(AOB) = \cot(180 - \theta) = -\cot \theta = -\frac{1}{\tan \theta} = -\frac{5}{12}](/tpl/images/1323/3619/9f42e.png)
Q2)
Plot (2,8) on the cartesian plane and name it 'A'Drop a perpendicular to x-axis from 'A' and name the point 'B' where the perpendicular meets x-axis.Join the point A with the origin 'O'.
You'll notice a right-angled ΔABO formed (∠B = 90°) in the 1st quadrant of the plane whose :-
length of perpendicular of triangle (AB) = 8 unitslength of base of triangle (OB) = 2 unitslength of hypotenuse (OA) = 2√17 units
Let the angle between OA & positive x-axis be θ.
⇒ ∠AOB = θ
So ,
![\sin(AOB) = \sin \theta = \frac{8}{2\sqrt{17} } = \frac{4}{\sqrt{17} }](/tpl/images/1323/3619/19642.png)
![\cos(AOB) = \cos \theta = \frac{2}{2\sqrt{17}} = \frac{1}{\sqrt{17}}](/tpl/images/1323/3619/23ae2.png)
![\tan(AOB) = \tan \theta = \frac{8}{2} = 4](/tpl/images/1323/3619/4915b.png)
![\csc(AOB) = \csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{17}}{4}](/tpl/images/1323/3619/dfa5c.png)
![\sec(AOB) = \sec \theta = \frac{1}{\cos \theta} = \sqrt{17}](/tpl/images/1323/3619/b72cc.png)
![\cot (AOB) = \cot \theta = \frac{1}{\tan \theta} = \frac{1}{4}](/tpl/images/1323/3619/6f567.png)
Q3)
Plot (3,-6) on the cartesian plane and name it 'A'Drop a perpendicular to x-axis from 'A' and name the point 'B' where the perpendicular meets x-axis.Join the point A with the origin 'O'.
You'll notice a right-angled ΔABO formed (∠B = 90°) in the 4th quadrant of the plane whose :-
length of perpendicular of triangle (AB) = 6 unitslength of base of triangle (OB) = 3 unitslength of hypotenuse (OA) = 3√5 units
Let the angle between OA & positive x-axis be θ . [Assume it in counterclockwise direction].
⇒ ∠AOB = 360 - θ
So ,
![\sin(AOB) = \sin(360 -\theta) = -\sin \theta = -\frac{6}{3\sqrt{5} } = -\frac{2}{\sqrt{5} }](/tpl/images/1323/3619/37e98.png)
![\cos(AOB) = \cos(360 - \theta) = \cos \theta = \frac{3}{3\sqrt{5} } = \frac{1}{\sqrt{5} }](/tpl/images/1323/3619/c702c.png)
![\tan(AOB) = \tan(360 - \theta) = -tan \theta = -\frac{6}{3} = -2](/tpl/images/1323/3619/d5af5.png)
![\csc(AOB) = \csc(360 - \theta) = -\csc \theta = -\frac{1}{\sin \theta} = -\frac{\sqrt{5} }{2}](/tpl/images/1323/3619/2ab90.png)
![\sec(AOB) =\sec (360 - \theta) = \sec \theta = \frac{1}{\cos \theta} = \sqrt{5}](/tpl/images/1323/3619/f9256.png)
![\cot(AOB) = \cot(360 - \theta) = -\cot \theta = -\frac{1}{\tan \theta} = -\frac{1}{2}](/tpl/images/1323/3619/2776e.png)