1)In a figure,O is the centre of the circle ,A,B,C and D are points on the circumference of the circle.A,O,X and C are points on a straight to the circle at D,angle BOC=48° and angle CAD=36°;
a)Determine the sizes of the following angles:i)OBD
ii)CED
b)Calculate the radius of the circle given that AB=7.6cm,correct to one decimal place
c)Given that X divides AC in the ratio 3:4,ED=12.9cm and EC=7.8cm,calculate the length of XC correct to one decimal place.
2)A sector of an angle is 108° is cut from a circle of a radius 21cm.It is folded to form a cone.Calculate to 1 decimal place: (use π=22/7);
a)The curved surface area of the cone
b)The base radius of the cone
c)The vertical height of the cone
d)A cone is chopped off this cone horizontally from the apex to form a frustum of height 12cm.Calculate the volume of the frustum.
3)Find the root to the equation;2x^2+5x-7=0 by completing the square method.
4)A rectangular hockey field measures 90 metres long by 55 metres wide.The dimensions are measured to the nearest 2 metres and 1 metre respectively. Calculate the exact maximum percentage error in the perimeter of the field.
5)Given that sin(3x-60°)+2cos(3x-60°)=Cos(3x-60°),solve for 0°<or=x<or=270°
6)Find the third proportional to √5,√10,leaving your answer in a simplified form.Do not use a calculator or mathematical tables.
7)Wambugu invested a sum of money in a financial institution that paid a compound interest at the rates of 6% and 7% respectively for the first two years.At the end of the second year, Wambugu withdrew Ksh 1701300.Determine the amount of money that he invested in the constitution.
a)
i) Since OC is the radius of the circle, angle BOC is equal to 48°. Therefore, angle OBC is half of angle BOC, so angle OBC = 24°.
ii) Angle CAD is 36°. Since AC is a straight line, angle CED is equal to angle CAD, so angle CED = 36°.
b) Since AB is a chord of the circle, OB is the perpendicular bisector of AB. Thus, OAB is a right-angled triangle. Using the Pythagorean theorem, we can calculate the radius of the circle:
(7.6/2)^2 + r^2 = 7.6^2
2.88 + r^2 = 57.76
r^2 = 54.88
r ≈ 7.4 cm
c) Let XC = 3x and AX = 4x. Using the property that the angle at the center of a circle is twice the angle at the circumference, we can calculate the angles:
angle ADC = angle AOC = 2 * angle CAD = 2 * 36° = 72°
angle BDC = 360° - angle BOC - angle ADC = 360° - 48° - 72° = 240°
Using the sine rule in triangle BDC, we have:
sin(72°) / 7.8 = sin(240°) / 4x
sin(72°) / 12.9 = sin(240°) / 3x
Solving these equations, we get x ≈ 2.6. Therefore, XC = 3x ≈ 7.8 cm.
2)
a) The curved surface area of the cone is equal to the length of the sector cut from the circle. The length of the sector is:
(108/360) * 2π * 21 = 22 cm
b) The base radius of the cone is equal to the radius of the original circle, which is 21 cm.
c) The vertical height of the cone can be calculated using the Pythagorean theorem. The slant height of the cone is equal to the radius of the original circle, which is 21 cm. The height of the cone is:
sqrt((21^2) - (21^2)) = 0 cm
d) Since the frustum is formed by chopping off the smaller cone from the larger cone, the volume of the frustum is equal to the volume of the larger cone minus the volume of the smaller cone. The volume of the smaller cone can be calculated using the formula:
1/3 * π * (12/21)^2 * 21 = 12π cm^3
The volume of the frustum is therefore:
1/3 * π * (21^2 + 21(12/21) + (12/21)^2) * 12 = 872π cm^3
3) To find the roots of the equation 2x^2 + 5x - 7 = 0 using the completing the square method, we can rewrite the equation as:
2(x^2 + (5/2)x) = 7
2(x^2 + (5/2)x + (5/4)^2) = 7 + 2(5/4)^2
2(x + (5/4))^2 = 7 + 25/8
2(x + (5/4))^2 = 63/8
Dividing both sides by 2, we get:
(x + (5/4))^2 = 63/16
Taking the square root of both sides, we have:
x + 5/4 = ±√(63/16)
x + 5/4 = ±√(63)/√(16)
x + 5/4 = ±3/4√(7)
x = (-5 ± 3√7)/4
So the roots of the equation are x = (-5 + 3√7)/4 and x = (-5 - 3√7)/4.
4) The perimeter of the field is given by P = 2(90 + 55) = 2(145) = 290 meters.
The maximum error in the length is 2 meters, and the maximum error in the width is 1 meter. Therefore, the maximum error in the perimeter is 2 + 2 + 1 + 1 = 6 meters.
The maximum percentage error is given by (6 / 290) * 100% = 2.1%.
5) By using the double angle formula for sin, we get:
sin(3x)cos(60°) + cos(3x)sin(60°) + 2cos(3x)cos(60°) = cos(3x)
Using the trigonometric identity sin(60°) = √3/2 and cos(60°) = 1/2, we have:
(√3/2)sin(3x) + (1/2)cos(3x) + (1/2)cos(3x) = cos(3x)
Combining like terms, we get:
(√3/2)sin(3x) + cos(3x) = cos(3x) - (1/2)cos(3x)
(√3/2)sin(3x) = (1/2)cos(3x)
Dividing both sides by cos(3x), we get:
(√3/2)tan(3x) = 1/2
Taking the inverse tangent of both sides, we have:
3x = arctan(1/(2√3))
x = arctan(1/(2√3))/3
Using a calculator, we can approximate this to x ≈ 13.2°.
6) The third proportional to √5 and √10 is √20, since (√5)^2/√10 = √20.
Therefore, the third proportional to √5 and √10 is √20.
7) Let the amount of money Wambugu invested be x.
After the first year, the amount of money becomes x + 0.06x = 1.06x.
After the second year, the amount of money becomes 1.06x + 0.07(1.06x) = 1.06x + 0.0742x = 1.1342x.
Given that he withdrew Ksh 1701300, we can set up the equation:
1.1342x = 1701300
Dividing both sides by 1.1342, we get:
x = 1701300 / 1.1342 ≈ 1500000
Therefore, Wambugu invested Ksh 1500000 in the constitution.