Right triangles and trigonometry #1 - Q&A

Question 1 Find an equation of the plane. The plane through the points (2,1,2), (3, -8, 6), and (-2, -3,1) Answer 1 Let the given points be P(2, 1, 2), Q(3, - 8, 6) and R(-2,-3.1) overline PQ = \langle3 - 2, - 8 - 1, 6 - 2\rangle = \langle1, - 9, 4\rangle PR = \langle- 2 - 2, - 3 - 1, 1 - 2\rangle = \langle- 4, - 4, - 1\rangle PQPR = [[1, - 9, 4], [- 4, - 4, - 1]] (9+16)+(-16+1)+(436) Therefore, the normal vector to the plane is tilde n = \langle25, - 15, - 40\rangle Since, the plane passes through all the three points we can choose any point to find its equation. So, the equation of the plane through the point P(2,1,2) with normal vector dot n = \langle25, - 15, - 40\rangle ; 25(x - 2) - 15(y - 1) - 40(z - 2) = 0 Rightarrow 25x - 50 - 15y + 15 - 40x + 80 = 0 25x - 15y - 40z + 45 = 0 Answer 2 To find an equation of the plane passing through the points (2,1,2), (3, -8, 6), and (-2,-3, 1), we can use the concept of cross products and the equation of a plane. Let's denote the three given points as P, (2.1.2), P(3.8.6), and P(-2.3.1). We need to find the normal vector of the plane, which can be obtained by taking the cross product of two vectors lying on the plane. Let's consider the vectors and defined as follows PP-8-1 PPP -3-(-5) Let's substitute the coordinates of one of the given points, such as P, (2.1.2), into the 3-2 -2-3 1-6 of the plane, we take the cross product of and To find the nor overline n = overline v 1 * overline v_{2} = [[1], [- 9], [4]] * [[- 5], [5], [- 5]] . expansion method, we can calculate the cross product as follows: -35 -15 (-9)(-5)-4(5) -4(-5)-1(-5) 1(5)-(-9)(-5) -35 Now, we have the normal vector -15 To find the equation of the plane, we can use 40 the point-normal form of the equation, given by: Ax+By+C=D where (A.B.C) is the normal vector and (xyz) represents any point on the plane. equation. This allows us to determine the value of D - 35(2) - 15(1) + 40(2) = D Simplifying, we have: - 70 - 15 + 80 = D -S-D Thus, the equation of the plane becomes: - 35x - 15y + 40z = - 5 Answer 3 Step 1. Find the vectors a and v using two of the given points: 3- 6-2 PR= -2-2 -3-1 1-2 Step 2. Calculate the cross product a=uxv = [[1], [- 9], [4]] * [[- 4], [- 4], [- 1]] = [[31], [- 15], [40]] Step 3. The equation of the plane can be written as Ax+ By+CzD. where a = is the normal vector to the plane. Substituting one of the given points, such as P(2.1.2), we can solve for D: 31(2) - 15(1) + 40(2) =Dm62-15+80=DmD=127. Step 4. Therefore, the equation of the plane is 11x - 15y + 40z = 127 Answer 4 Result - 52x + 15y - 40z + 169 = 0 Solution: Let's assume the equation of the plane is Ax+By+Cz+D0, where A, B, C are the coefficients of the variables x, y, z, respectively, and D is a constant. First, we need to find the normal vector to the plane. We can obtain the normal vector by taking the cross product of two vectors in the plane. We can choose two vectors as follows: vec v_{1} = (3 - 2 - 8 - 1.6 - 2) = (1, - 9, 4) vec v_{2} = (- 2 - 2 - 3 - 1, 1 - 2)(- 4, - 4, - 1) To find the normal vector, we calculate the cross product: -94 Expanding the determinant, we have: N=(-36-16):-(1-16)-(4+36)-521+157-40 Now, we can use one of the given points (let's choose (2,1,2)) to find the value of D in the equation of the plane. Substituting the coordinates of the point into the equation, we have: - 52(2) + 15(1) - 40(2) + D = 0 - 169 + D = 0 D = 169 Therefore, the equation of the plane is - 52x + 15y - 40z + 169 = 0 Question 2 Wryin * cos(- pi/6) the same as cos()? Answer 1 Cosine has an even function. This means that cos(- pi/6) * i the same thing as cos(pi/6) Answer 2 cos(- x) = cos(x) the cos function is even. Question 3 What is the value of x? Enter your answer in the box. x= X 24 7 Answer 1 By the Pythagorean theorem (for right triangles) : c ^ 2 = a ^ 2 + b ^ 2 where a, b, and c are the lengths of the legs and the hypotenuse, respectively , a = 7 ,b=24.c=x. x ^ 2 = 7 ^ 2 + 24 ^ 2 = 49 + 576 = 625 Next, we obtain the value of x by taking the square root: sqrt(pi ^ 2) = sqrt(6) * 25 = 25 x = 25 Question 4 Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. Answer 1 Step From the figure, by Pythagoras theorem we can write sigma/2 = sqrt(r ^ 2 - x ^ 2) Step 2 Area of Triangle (Base) (Height) 2 A = 1/2 * (b)(r + x) Use Eqn(1) to substitute the value of b A = 1/2 * (2sqrt(r ^ 2 - x ^ 2))(r + x) A(x) = (r + x) * sqrt(r ^ 2 - x ^ 2) Step Differentiate A(z) A' * (x) = d/dx [(r + x) * sqrt(r ^ 2 - x ^ 2)] A' * (x) = sqrt(r ^ 2 - x ^ 2) * d/dx [r + x] + (r + x) * d/dx [sqrt(r ^ 2 - x ^ 2)] A^ prime (x)= sqrt r^ 2 -x^ 2 *1+(r+x)* d[ sqrt r^ 2 -x^ 2 ] d(r ^ 2 - x ^ 2) * (d(r ^ 2 - x ^ 2))/(dx) A' * (x) = sqrt(r ^ 2 - x ^ 2) + (r + x) * 1/(2sqrt(r ^ 2 - x ^ 2)) * (- 2x) A' * (x) = sqrt(r ^ 2 - x ^ 2) - (rx + x ^ 2)/(sqrt(r ^ 2 - x ^ 2)) Solve for A' * (x) = 0 sqrt(r ^ 2 - x ^ 2) - (rx + x ^ 2)/(sqrt(r ^ 2 - x ^ 2)) = 0 sqrt(r ^ 2 - z) Answer 2 Step 1 The equation of circle will be x ^ 2 + y ^ 2 = r ^ 2 Step 2 equating to zero (r ^ 2 - ry - 2y ^ 2)/- 2 = 0 v' - y 2 - pi*v - 2v ^ 2 = 0 r ^ 2 - ry - 2y ^ 2 = 0 = (r plus/minus sqrt(9r ^ 2))/4 = rejecting, y Therefore, yor we check to see A'()>0 for y < r/2 and therefore y = r/2 maximum y = r/2 y = r/2 maximum A' * (y) < 0toty > tau/2 x^ 2 +( r 2 )^ 2 Rightarrow r^ 2 Rightarrow x = (sqrt(3))/2 * r Dimension of triangle with maximum area is base = 2x = 2 * (sqrt(3))/2 * r = sqrt(3) * r h = r + y = r + r/2 = 3/2 * r Question 5 How do you estimate an angle to the nearest one-half radian? Answer 1 Start with drawing of trigonometric circle. Locate the four main rays of the trigonometric circle. 3 angles 0. Also, locate rays that represent angles 3 5 7 It is now easy to roughly approximation an angle to the nearest ray if you located these rays For example, if the angle is between rays that represent and we round that angle to