Mathematics, 14.05.2021 04:50 sulemmaa

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

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Answer from: alejandro1102

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 ,

$\sin (AOB) = \sin(180 - \theta) = \sin \theta = \frac{12}{13}$ $\cos(AOB) = \cos (180 - \theta) = -\cos \theta = -\frac{5}{13}$ $\tan(AOB) = \tan(180 - \theta) = -tan \theta = -\frac{12}{5}$ $\csc(AOB) = \csc(180 - \theta) = \csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}$ $\sec(AOB) = \sec(180 - \theta) = -\sec \theta = -\frac{1}{\cos \theta} = -\frac{13}{5}$ $\cot(AOB) = \cot(180 - \theta) = -\cot \theta = -\frac{1}{\tan \theta} = -\frac{5}{12}$

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 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 ,

$\sin(AOB) = \sin \theta = \frac{8}{2\sqrt{17} } = \frac{4}{\sqrt{17} }$ $\cos(AOB) = \cos \theta = \frac{2}{2\sqrt{17}} = \frac{1}{\sqrt{17}}$ $\tan(AOB) = \tan \theta = \frac{8}{2} = 4$ $\csc(AOB) = \csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{17}}{4}$ $\sec(AOB) = \sec \theta = \frac{1}{\cos \theta} = \sqrt{17}$ $\cot (AOB) = \cot \theta = \frac{1}{\tan \theta} = \frac{1}{4}$

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 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 ,

$\sin(AOB) = \sin(360 -\theta) = -\sin \theta = -\frac{6}{3\sqrt{5} } = -\frac{2}{\sqrt{5} }$ $\cos(AOB) = \cos(360 - \theta) = \cos \theta = \frac{3}{3\sqrt{5} } = \frac{1}{\sqrt{5} }$ $\tan(AOB) = \tan(360 - \theta) = -tan \theta = -\frac{6}{3} = -2$ $\csc(AOB) = \csc(360 - \theta) = -\csc \theta = -\frac{1}{\sin \theta} = -\frac{\sqrt{5} }{2}$ $\sec(AOB) =\sec (360 - \theta) = \sec \theta = \frac{1}{\cos \theta} = \sqrt{5}$ $\cot(AOB) = \cot(360 - \theta) = -\cot \theta = -\frac{1}{\tan \theta} = -\frac{1}{2}$

Answer from: Quest

idk

step-by-step explanation:

do it yourself, cause that made me do it myself today, in english!

Answer from: Quest

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!

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