Find the values of x such that the angle between the vectors <2, 1, -1> and <1, x, 0> is 45?
cos(theta) = a.b / (|a|*|b|). Let a = <2,1,-1> , b = <1,x,0> So we first take the dot product of the two: a.b = 2 + x. Then we find the magnitudes of the two vectors. |a| = sqrt (6), |b| = sqrt(1+x^2) Now for the angle between the two vectors to be 45 degrees, we substitute 45 degrees into cosx and try to find x. cos(45) = 1/sqrt(2) Now we substitute everything back into the first equation: 1/sqrt(2) = (2+x) / sqrt(6+6x^2) So sqrt(6+6x^2) = (2+x) * sqrt(2). sqrt(3+3x^2) = 2+x (I divided by sqrt(2) on both sides) 3+3x^2 = 4+4x+x^2 (square both sides) 2x^2 -4x-1 = 0 which gives you x = 1 + (sqrt(6)/2), x = 1 - (sqrt(6)/2)
Find the values of x such that the angle between the vectors (2,1,-1) and (1,x,0) is $5 degrees?
Dot product equation is: a·b = |a||b| cosθ Evaluating each part separately: a·b = (2*1)+(1*x)+(-1*0) = x+2 |a| = √[2²+1²+(-1)²] = √[4+1+1] = √6 |b| = √[1²+x²+0²] = √[x²+1] cosθ = 1/√2. Putting the equation together: x+2 = √6√[x²+1]/√2. To solve, square the equation: x²+4x+4 = 6*(x²+1)/2; x²+4x+4 = 3x² + 3; 2x²-4x-1 = 0. Using the quadratic formula, x = (4±2√6)/4. Both answers check out if substituted back into the original dot product equation (SOP for solving equations with radicals).
If sin(x)=3/4, and x is an acute angle, show the exact value of cos(x) (do not find the value of x)?
Draw a right triangle in the first quadrant as angle x is acute and both sin and cos are positive. Since sin(x) = o/h = 3/4, the side opposite the angle could be considered 3 and the hypotenuse 4. Since you need the adjacent side for cos, find the other side by pythagorean theorem, which is sqrt(7). Then cos = a/h = sqrt(7)/4.
Angle ABD and DBC are supplementary, find the value of x.?
if they are supplementary, then they are 180 degrees total. so, 14x (ABD) + 8x+4 (DBC) = 180 now solve for x 14x+8x+4=180 22x+4=180 -4 -4 22x=176 /22 /22 x=8 also, if you want to know, if you plug in x to the angle values you get 14(8)=112 8(8)+4=68 angle ABD=112 and angle ABD=68 if you are in geometry, you will need to know how to do this too
What is the value of angle [math]x[/math] in the triangle shown in the details?
AB = AC -> 70 + a = 60 + 20 -> a = 10ABC + BCA + CAB = 180 -> 80 + 80 + b = 180 -> b = 20[BOC -> 180 - 70 - 60 = 50, thus DOE = 50, BOD = COE = 130][BDO -> 180 - 130 - 10 = 40, CEO -> 180 - 130 - 20 = 30[if you put x = OED = 60, ODE -> 180 - 60 - 50 = 70, ADE = 70 and AED = 90]SimilarlyThus AED is right angle triangle at E and is congruent with CED with x = 60
Find the angle made by the line joining (5,3) and (-1,-3) with the positive direction of x?
The points joining the line are (5,3) and (-1,-3).Let's find the slope of the line.[math]m = \dfrac{\Delta y}{\Delta x}[/math]Or,[math]m = \dfrac{3-(-3)}{5-(-1)}[/math][math]m = \dfrac{6}{6}[/math]Or, [math]m = 1[/math]Now, we know that the slope of a straight line is the tangent of the angle made between the x-axis and the line.Let this angle be [math]\theta[/math]Hence,[math]tan \theta = m[/math][math]tan \theta = 1[/math]But,[math]tan 45° = 1[/math]Hence,[math]\theta = 45°[/math]
The lengths of the sides of a triangle are 4-x, x+2 and 2x-2. Find all values of x for which the triangle is isosceles?
The first thing to remember when attempting to construct any triangle is that the sum of the lengths of any two sides MUST be greater than the length of the other side. To see why, just picture a side of arbitrary length with the two other sides on hinges. If the two other sides do not add up to at least as long as that side, no matter what angle you put them at, they will never touch, and if they are they add up to the same as that side, all three sides would be co-linear, which makes a line, not a triangle. So if we can’t have any two sides add up to more than the other side, we are stuck, and will never get a triangle.With that said, let’s consider how the two sides with lengths [math]4-x[/math] and [math]2x-2[/math] compare to the length of the other side ([math]x+2[/math]):[math](4-x)+(2x-2)=(2x-x)+(4-2)=x+2[/math]Uh oh. We are in trouble. These two sides add up to exactly the same length as the other side, no matter what [math]x[/math] is. So the given lengths can never create a triangle. They will always just provide a line segment of length [math]x+2[/math]. There is no solution to this problem.
If [math]\sin x = \cos x[/math] and [math]x[/math] is acute, what is the value of [math]x[/math]?
The value of sin and cos coincide at half of every quadrant, though the signs vary accordingly.Now since you asked acute angle for the above condition, the only value is 45° or (pi)/4.And the value is 1/√2.