trig functions (1 Viewer)

[tradewind]

i need help with these.

* if the area of a circle is 200cm^2 amd a sector is cut off be an angle of 3pi /4 at the centre, find the area of the sector?

* The area of a sector is 3pi /10 cm^2 and the arc length cutt off by the sector is pi /5 cm. Find the angle subtended at the centre of the circle and find the radius of the circle?

* the length of an arc is 8.9 cm and the area of the sector is 24.3 cm ^2 when an angle of x is subtended at the centre of a ciecle. Find the area of the minor segment cut off by x , correct to one decimal place

* find the diameter of the sun to the nearest km if its distance from the earth is 149 000 000 km and it subtends at an angle of 31' at the earth

*Given that the wingspan of an aeroplane is 30 m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' when it is diectly overhead

 

[CM_Tutor]

Originally posted by tradewind
i need help with these.

* if the area of a circle is 200cm^2 amd a sector is cut off be an angle of 3pi /4 at the centre, find the area of the sector?

Use A = pi * r² with A = 200 cm² to find r, then use that r value and @ = 3 * pi / 4 in A_sector = r²@ / 2

* The area of a sector is 3pi /10 cm^2 and the arc length cutt off by the sector is pi /5 cm. Find the angle subtended at the centre of the circle and find the radius of the circle?

A = r²@ / 2, where A = 3 * pi / 10 cm² and l = r@ where l = pi / 5 cm - this produces two simultaneous equations in two unknowns, r and @, which you need to solve.

* the length of an arc is 8.9 cm and the area of the sector is 24.3 cm ^2 when an angle of x is subtended at the centre of a ciecle. Find the area of the minor segment cut off by x , correct to one decimal place

Use the method of the previous question to find r and @, and then note that
A_{minor segment} = x = (1 / 2) * r²(@ - sin @).

* find the diameter of the sun to the nearest km if its distance from the earth is 149 000 000 km and it subtends at an angle of 31' at the earth

Draw a diagram of an isosceles triangle. The apex angle (on Earth) is 31', and the base is a chord on the sun. The perpendicular bisector of the chord passes through the centre of the sun, and bisects the apex angle. Join the radii from the centre of the sun to the ends of the base of the isosceles triangle. We now have a right angled triangle in which tan (31' / 2) = R / 149 000 000, where R is the radius of the sun in kilometres.

*Given that the wingspan of an aeroplane is 30 m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' when it is diectly overhead

This uses the identical method to the sun question above.

 

[tradewind]

thanks
btw how do i do this

find all the solutions to the following equations in 0 < x< 2pi

* cos2x = 1/2

 

[Courtney]

Ok, I hope this is right...

cos 2x=1/2
2x=60 (inverse cos of 1/2)
x=30

in terms of pi
x=pi/6

noting cos is positive in 1st and 4th quadrant only
x=pi/6, (2pi-pi/6)
x=pi/6, 7pi/6

I hope that helps..

 

[CM_Tutor]

Courtney, you have missed two solutions, and 2 * pi - pi / 6 does not equal 7 * pi / 6

Answer:

We seek to solve cos2x = 1 / 2 for 0 <= x <= 2* pi
ie for 0 <= 2x <= 4 * pi.

The related angle is pi / 3, as cos(pi / 3) = 1 / 2, and we want angles (2x) to be in the 1st and 4th quadrants.

So, 2x = pi / 3, 2 * pi - pi / 3, 2 * pi + pi / 3, or 4 * pi - pi / 3
2x = pi / 3, 5 * pi / 3, 7 * pi / 3, or 11 * pi / 3

So, x = pi / 6, 5 * pi / 6, 7 * pi / 6 or 11 * pi / 6

 

[Estel]

A very long winded way but effective in ensuring you don't miss solutions is to do a 2 second sketch.