1)In a figure,ABC is a tangent at B and CDE is a straight line.Angle BED=50°,angle DEF=35° and angle ECB=25°.Calculate the values of x and z
2)The equation of a circle is given by 4x^2+4y^2-8x+2y-7=0.Determine the coordinates of the centre of the circle
3)A sequence us formed by adding corresponding terms of an AP and GP.The first, second and third terms of the sequence formed are 14,34 and 78 respectively.
a)Given that the common ratio of the G.P is 3;
i)Find the first term of the AP and GP and the common difference of the AP
ii)Find the sixth term and the sum of the first six terms of the sequence
b)The second and third terms of a geometric progression are 24 and 12(x+1)respectively. Find the whole number value of x and hence the first term given the sum of the first three terms of the progression is 76.
4)A diagram shows a sketch of line y=3x and the curve y=4-x^2 intersecting at point P and Q.
a)Find the coordinates of P and Q
b)Given that QN is perpendicular to the x-axis at N,calculate;
i)The area bounded by the curve y=4-x^2,the x-axis and line QN
ii)The area of the shaded region that lies below the x-axis.
iii)The area of the region enclosed by the curve y=4-x^2,the line y=3x and the y-axis
5)The position of two towns A and B are (50°N,45°W) and (50°N,K°W) respectively.It takes a plane 5 hours to travel from A to B at an average speed of 800knots.The same plane takes 1 and 1/2 hours to travel from B to another town C at the same average speed.Given that C is to the north of B,calculate to the nearest degree;
a)The value of K
b)The latitude of C
1) Let's analyze each part of the figure separately:
a) In triangle BED, we have:
Angle BED = 50°
Angle EDB = 180° - BED - DEF = 180° - 50° - 35° = 95°
Since ABC is a tangent, angle ECB = angle EDB = 95°
In triangle CEB:
Angle ECB = 25°
Angle CBE = 180° - ECB - EBC = 180° - 25° - 90° = 65°
Now, we know that in triangle CDE:
Angle DCE = 180° - CBE = 180° - 65° = 115°
Since DE is a straight line, angle DEC = 180° - 115° = 65°
Finally, in triangle DFE:
Angle DEF = 35°
Angle DFE = 180° - DEF - DEC = 180° - 35° - 65° = 80°
Thus, x = angle CDE = 115° and z = angle DFE = 80°
2) To determine the center of the circle, we need to complete the square for x and y in the given equation:
4x^2 - 8x + 4 + 4y^2 + 2y + 1 - 4 - 1 - 7 = 0
(2x - 2)^2 + (2y + 1)^2 - 12 = 0
Comparing this with the standard form for a circle, (x-a)^2 + (y-b)^2 = r^2, we find:
Center of the circle: (a, b) = (1, -1)
Radius of the circle: r^2 = 12
3)
a)
i) Let the first term of the AP be 'a' and the common difference be 'd'. Then:
a + 14 = 14
a + 34 = a*3
a + 78 = a*3^2
Solving these equations, we find:
a = 2, d = 12
ii) The sixth term of the sequence is:
2 + 14 + 34 + 78 + 2*3 + (2+12)*3 = 2 + 14 + 34 + 78 + 6 + 42 = 176
Sum of the first six terms:
6/2(2 + 176) = 6*178/2 = 534
b) Given that the second term is 24 and the third term is 12(x+1), we can solve for x:
24*r = 24
24*r^2 = 12(x+1)
Solving these equations, we find:
r = 1, x = 1
Therefore, the first term of the progression is:
24/r = 24
4)
a) To find the coordinates of P and Q, we need to solve the simultaneous equations y = 3x and y = 4 - x^2:
3x = 4 - x^2
x^2 + 3x - 4 = 0
(x + 4)(x - 1) = 0
So, the coordinates of P and Q are:
P(-4, 12) and Q(1, 3)
b)
i) Area bounded by the curve y=4-x^2, the x-axis, and line QN:
Integrate the curve from -4 to 1: ∫(4-x^2)dx from -4 to 1 = [4x - (x^3)/3] from -4 to 1 = (16 - 1/3) = 47/3 sq units
ii) Area of the shaded region below the x-axis is the negative of the integral: -47/3 sq units
iii) Area enclosed by the curve y=4-x^2, line y=3x, and the y-axis:
Integrate the curve from -4 to 1 and from 0 to 3: (3 - x)dx from 0 to 3 = 27/2 sq units
5)
a)
Given that the distance from A to B is 800 knots * 5 hours = 4000 knots, we can use the cosine rule to find K:
cos(K) = (AB^2 + AC^2 - BC^2) / (2*AB*AC)
cos(K) = (4000^2 + 4000^2 - 4000^2) / (2*4000*4000)
cos(K) = 1/2
K = 60°
b)
Since C is to the north of B, the latitude of C is 50°N.