Question

I need help with the solutions for 19,20,21 thank you

Answer 1

GIVEN :-

Coordinates of points are :-

1. (-5 , 12)
2. (2 , 8)
3. (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 units
• length of base of triangle (OB) = 5 units
• length 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 = (12)/(13)`
• 
  `\cos(AOB) = \cos (180 - \theta) = -\cos \theta = -(5)/(13)`
• 
  `\tan(AOB) = \tan(180 - \theta) = -tan \theta = -(12)/(5)`
• 
  `\csc(AOB) = \csc(180 - \theta) = \csc \theta = (1)/(\sin \theta) = (13)/(12)`
• 
  `\sec(AOB) = \sec(180 - \theta) = -\sec \theta = -(1)/(\cos \theta) = -(13)/(5)`
• 
  `\cot(AOB) = \cot(180 - \theta) = -\cot \theta = -(1)/(\tan \theta) = -(5)/(12)`

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 units
• length of base of triangle (OB) = 2 units
• length of hypotenuse (OA) = 2√17 units

Let the angle between OA & positive x-axis be θ.

⇒ ∠AOB = θ

So ,

• 
  `\sin(AOB) = \sin \theta = (8)/(2√(17) ) = (4)/(√(17) )`
• 
  `\cos(AOB) = \cos \theta = (2)/(2√(17)) = (1)/(√(17))`
• 
  `\tan(AOB) = \tan \theta = (8)/(2) = 4`
• 
  `\csc(AOB) = \csc \theta = (1)/(\sin \theta) = (√(17))/(4)`
• 
  `\sec(AOB) = \sec \theta = (1)/(\cos \theta) = √(17)`
• 
  `\cot (AOB) = \cot \theta = (1)/(\tan \theta) = (1)/(4)`

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 units
• length of base of triangle (OB) = 3 units
• length 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 = -(6)/(3√(5) ) = -(2)/(√(5) )`
• 
  `\cos(AOB) = \cos(360 - \theta) = \cos \theta = (3)/(3√(5) ) = (1)/(√(5) )`
• 
  `\tan(AOB) = \tan(360 - \theta) = -tan \theta = -(6)/(3) = -2`
• 
  `\csc(AOB) = \csc(360 - \theta) = -\csc \theta = -(1)/(\sin \theta) = -(√(5) )/(2)`
• 
  `\sec(AOB) =\sec (360 - \theta) = \sec \theta = (1)/(\cos \theta) = √(5)`
• 
  `\cot(AOB) = \cot(360 - \theta) = -\cot \theta = -(1)/(\tan \theta) = -(1)/(2)`