trig question (1 Viewer)

[pLuvia]

1. A rectangular strip of metal 10cm wide is bent to form a water channel, the section perpendicular to the length being a circular arc whose chord is 6cm long. Show that, if the circular arc subtends an angle of 2x radians at the centre of the circle,

5sinx=3x

Solve this equation graphically to three decimal places and find the area of the cross-section of the channel in cm²

2. An arc AB of a circle, centre C is 20cm long. If the chord AB is 15 cm long and subtends an angle of x radians at C, show that 8sin (x/2) = 3x. Find, graphically, the solution to this equation

 

[Mountain.Dew]

1. A rectangular strip of metal 10cm wide is bent to form a water channel, the section perpendicular to the length being a circular arc whose chord is 6cm long. Show that, if the circular arc subtends an angle of 2x radians at the centre of the circle,

5sinx=3x

Solve this equation graphically to three decimal places and find the area of the cross-section of the channel in cm²

2. An arc AB of a circle, centre C is 20cm long. If the chord AB is 15 cm long and subtends an angle of x radians at C, show that 8sin (x/2) = 3x. Find, graphically, the solution to this equation

ill do question 2...

here goes, my 2 cents.

using l = r@, @ = x, l = 20 so r = 20/x

we have triangle ABC, AB = 15, BC = AC = r = 20/x, and ^ACB = x

now, since ABC is isoceles triangle, ^BAC = ^ABC = 90 - (x/2)

therefore, using sine rule:

sin(^ACB) / 15 = sin^(BAC) / (20/x)

from here, substitute, get trig ratios in terms of angles of x/2, do a bit of simplifying, and voila!

8sin (x/2) = 3x

the solutions part i have no idea with...unless you somehow graph it on a number plane...

Q1 is very similar to Q2. apply the same principles.

 

[pLuvia]

Yeh, the solutions bit you have to graph which is really annoying thanks again M.D

 

[Mountain.Dew]

np glad to help.