1) sin(4x-10°) - cos(x+60°) = 0
sin(4x-10°) = cos(x+60°)
sin(4x-10°) = sin(30°-x)
4x-10° = 30° - x (since both angles have the same sine)
5x = 40°
x = 8°
2) a) sin A = √(1-cos^2A) = √(1-0.64^2) ≈ 0.768
b) Tan(90-A) = 1/TanA = 1/√(1-cos^2A)/cosA = cosA/√(1-cos^2A) = 0.8/0.48 ≈ 1.667
3) We can form a right triangle with the height of the building as the hypotenuse, and the distances from Paul and Omondi to the building as the adjacent legs. Let h be the height of the building in meters, then we have:
h/tan70° + 10/tan46.8° = h
Solving for h:
h = (10/tan46.8°)/(1/tan70° - 1)
h ≈ 20.8 meters
4) 1/2(24-4x) > 6(3x-4/3) ≥ -2/3(42+3x)
12-2x > 18x - 8 ≥ -28 - 2x
14x < 20 < 20x
1 < x < 5/7
The integral solutions are x=1, 2, 3, 4.
5) a) x-5 ≤ 3x-8 < 2x-3
-x ≤ -3 < -x+2
-5 ≤ x < -2
b) On a number line, we can represent the solution as a closed interval between the endpoints -5 and -2, with a filled-in circle at -5 and an open circle at -2.
1)solve for x in the equation :sin(4x-10°)-cos(x+60°)=0
2)Given that cosA=0.64,without using mathematical table or calculator find;
a)sin A
b)Tan (90-A)
3)Two students Paul and Omondi standing 10m apart on the same side of a tall building on a horizontal ground.Paul who is closer to the building sees the roof top at an angle of 70° while Omondi at an angle of 46.8° if the building Paul and Omondi lies on a straight line calculate the height of the building to 3 significant figures
4)solve the following inequalities and state the integral solutions;1/2(24-4x)>6(3x-4/3)>\=-2/3(42+3x)
5)Given that the inequalities x-5<\=3x-8<2x-3;
a)solve the inequalities
b) represent the solution on a number line.
1 answer
1 answer