1)find the values of x in sin60degrees=cos(3x-45)degrees
2)solve the equation sin(4x-10)degrees-cos(x+60)degrees=0
3)two students paul and omondi standing 10m apart on the same side of a tall building on a horizontal gruond .Paul who is closer to the building sees the roof top at an angle of 70degrees while omondi at ban angle of46.8degrees if the building ,paul and omondi lies on astraight line ,calculate the height of the building correct to 3significant figures
4)when a shirt is sold at 126shilling a loss of x%is made if the same shirt is sold at 154shilling a profit of x%is realized find the buying price of the shirt
1) sin 60° = cos(3x-45)°
Sin 60° = cos(45-(3x-45))° [using cos(A-B) = cosAcosB + sinAsinB]
Sin 60° = cos(90-3x)°
Sin 60° = sin3x°
3x = 60° + n(360°) or 180° - 60° + n(360°) [using sinθ = sin(180-θ)]
x = 20° + n(120°) or 40° + n(120°)
2) sin(4x-10)° - cos(x+60)° = 0
sin(4x-10)° = cos(30-(x+60))° [using cos(A-B) = cosAcosB + sinAsinB]
sin(4x-10)° = sin(x-30)°
4x-10 = n(360°) + (x-30)° or 180° - (x-30)° + n(360°) [using sinθ = sin(180-θ)]
3x = 40° + n(360°) or 150° + n(360°)
x = 40/3° + n(120°) or 50° + n(120°)
3) Let the height of the building be h.
From Paul's position, tan70° = h/x where x is the distance between Paul and the building.
From Omondi's position, tan46.8° = h/(x+10)
Since Paul and Omondi lie on the same straight line, we have x + h/tan70° = (x+10) + h/tan46.8°
Solving for h, we get h = 15.676m (approx)
4) Let the buying price of the shirt be y shillings.
Selling price at a loss of x% = y - (xy/100)
Selling price at a profit of x% = y + (xy/100)
Given, y - (xy/100) = 126 and y + (xy/100) = 154
Solving for y, we get y = 140 shillings.
Therefore, the buying price of the shirt is 140 shillings.
1) sin 60° = cos(3x-45)°
cos30° = cos(3x-45)°
3x-45 = 30
x = 25
There is another solution in QIV, where cosine > 0
cos(3x-45)° is positive in the second and fourth quadrants. In the second quadrant, we have:
sin 60° = cos(3x-45)°
cos(180-60)° = cos(3x-45)° [using cos(180-θ) = -cosθ]
cos120° = cos(3x-45)°
3x-45 = 240 [using cosθ = cos(360-θ)]
3x = 285
x = 95
In the fourth quadrant, we have:
sin 60° = cos(3x-45)°
cos(360-60)° = cos(3x-45)° [using cos(360-θ) = cosθ]
cos300° = cos(3x-45)°
3x-45 = 420 [using cosθ = cos(360+θ)]
3x = 465
x = 155
So, the solutions for x are x = 25°, 95°, and 155°.
1) To find the values of x in the equation sin(60 degrees) = cos(3x-45 degrees), we can use the trigonometric identities. First, let's simplify the equation using the identities:
sin(60 degrees) = cos(3x-45 degrees)
√3/2 = cos(3x-45 degrees)
Now, we need to find the values of x that satisfy this equation. To do so, we'll solve the equation for cos(3x-45 degrees) and then find the corresponding values of x.
cos(3x-45 degrees) = √3/2
Using the inverse cosine function, we can find the principal value of (3x-45 degrees):
3x-45 degrees = arccos(√3/2)
Now, we need to solve for x:
3x = arccos(√3/2) + 45 degrees
x = (arccos(√3/2) + 45 degrees) / 3
By plugging in the value of arccos(√3/2) ≈ 30 degrees, we can find the values of x:
x = (30 degrees + 45 degrees) / 3
x = 75 degrees / 3
x = 25 degrees
Therefore, one possible solution for x is 25 degrees.
2) To solve the equation sin(4x-10 degrees) - cos(x+60 degrees) = 0, we'll use the same approach as the previous question. Let's begin by simplifying the equation:
sin(4x-10 degrees) - cos(x+60 degrees) = 0
Next, let's solve for sin(4x-10 degrees):
sin(4x-10 degrees) = cos(x+60 degrees)
To find the values of x that satisfy this equation, we'll equate the trigonometric functions using the identities:
sin(4x-10 degrees) = sin(90 degrees - (x+60 degrees))
Now, we can set up the equation:
4x - 10 degrees = 90 degrees - (x+60 degrees)
Simplifying further:
4x - 10 degrees = 90 degrees - x - 60 degrees
4x + x = 90 degrees + 60 degrees - 10 degrees
5x = 140 degrees
x = 140 degrees / 5
x = 28 degrees
Therefore, one possible solution for x is 28 degrees.
3) To calculate the height of the building in this scenario, we can use the concept of similar triangles. Let's denote the height of the building as h.
From the given information, we have two right triangles formed by the students and the building. Let's consider Paul's triangle:
We know that Paul sees the rooftop at an angle of 70 degrees, and the distance between Paul and the building is 10m. Therefore, using the tangent function:
tan(70 degrees) = h / 10m
Solving for h:
h = 10m * tan(70 degrees)
Using a calculator, we find that tan(70 degrees) is approximately 2.747.
h ≈ 10m * 2.747
h ≈ 27.47m
Therefore, the height of the building is approximately 27.47m.
4) In this problem, let's assume the buying price of the shirt is B shillings.
Given that the shirt is sold at a loss of x% for 126 shillings, we can set up the equation:
B - (x/100)B = 126
Simplifying further:
B - (x/100)B = 126
B(1 - x/100) = 126
B = 126 / (1 - x/100)
Now, given that the shirt is sold at a profit of x% for 154 shillings, we can set up another equation:
B + (x/100)B = 154
Simplifying further:
B + (x/100)B = 154
B(1 + x/100) = 154
B = 154 / (1 + x/100)
Since the buying price of the shirt (B) is the same in both equations, we can equate them:
126 / (1 - x/100) = 154 / (1 + x/100)
Cross-multiplying:
126(1 + x/100) = 154(1 - x/100)
Expanding:
126 + 1.26x = 154 - 1.54x
Combining like terms:
1.26x + 1.54x = 154 - 126
2.8x = 28
x = 28 / 2.8
x = 10
Now that we have the value of x, we can substitute it back into either equation to find the buying price of the shirt:
B = 126 / (1 - 10/100)
B = 126 / (1 - 0.1)
B = 126 / 0.9
B = 140
Therefore, the buying price of the shirt is 140 shillings.