Determine the third angle of the triangular apparatus.

Here are the solutions to your tasks: a) Help John to know the length of the remaining sides and angle of the remaining vertex. Step 1: Determine the third angle of the triangular apparatus. The sum of angles in a triangle is 180^. Given angles are 59^ and 73^. Remaining angle = 180^ - 59^ - 73^ Remaining angle = 180^ - 132^ Remaining angle = 48^ The angle of the remaining vertex is 48^. Step 2: Use the Sine Rule to find the lengths of the remaining sides. Let the given side be c = 12 ft, opposite the 48^ angle. Let the other angles be A = 59^ and B = 73^, with opposite sides a and b respectively. The Sine Rule states: (a)/( A) = (b)/( B) = (c)/( C). We have (a)/( 59^) = (b)/( 73^) = (12)/( 48^). Step 3: Calculate side a. a = (12 59^)/( 48^) a ≈ (12 × 0.8572)/(0.7431) a ≈ (10.2864)/(0.7431) a ≈ 13.84 ft The length of one remaining side is 13.84 ft. Step 4: Calculate side b. b = (12 73^)/( 48^) b ≈ (12 × 0.9563)/(0.7431) b ≈ (11.4756)/(0.7431) b ≈ 15.44 ft The length of the other remaining side is 15.44 ft. b) Guide the learners on how to come up with the angle in the experiment present them in a solution set. Step 1: Rewrite the given equation using trigonometric identities. The equation is 5 - 2 ^2 = ^2 - 1. Recall the identity ^2 = 1 + ^2 , which implies ^2 = ^2 - 1. Substitute ^2 - 1 with (^2 - 1) - 1: 5 - 2 ^2 = (^2 - 1) - 1 5 - 2 ^2 = ^2 - 2 Step 2: Rearrange the equation into a quadratic form. Move all terms to one side to form a quadratic equation in terms of : 0 = ^2 + 2 ^2 - 5 - 2 3 ^2 - 5 - 2 = 0 Step 3: Solve the quadratic equation for . Let x = . The equation becomes 3x^2 - 5x - 2 = 0. Factor the quadratic equation: (3x + 1)(x - 2) = 0 This gives two possible values for x: 3x + 1 = 0 x = -(1)/(3) x - 2 = 0 x = 2 So, = -(1)/(3) or = 2. Step 4: Convert to and find the angles in the range 0^ 360^. Since = (1)/( ), we have = (1)/( ). Case 1: = (1)/(-1/3) = -3. This is not possible because the range of is [-1, 1]. So, there are no solutions from this case. Case 2: = (1)/(2). The principal value (reference angle) for = (1)/(2) is = ^-1((1)/(2)) = 60^. Since is positive, lies in Quadrant I and Quadrant IV. In Quadrant I: _1 = = 60^. In Quadrant IV: _2 = 360^ - = 360^ - 60^ = 300^. The solution set for is \60^, 300^\. c) Take the students through the steps on how B = (1 - 2 A - ^2 A)/(1 + 2 A - ^2 A) holds. Given that B = 45^ - 2A. We need to show that B equals the given expression. Step 1: Apply the tangent subtraction formula. The formula for (X - Y) is (X - Y) = ( X - Y)/(1 + X Y). Substitute X = 45^ and Y = 2A: B = (45^ - 2A) = ( 45^ - 2A)/(1 + 45^ 2A) Step 2: Substitute the value of 45^. We know that 45^ = 1. B = (1 - 2A)/(1 + 1 · 2A) B = (1 - 2A)/(1 + 2A) Step 3: Apply the tangent double angle formula. The formula for 2A is 2A = (2 A)/(1 - ^2 A). Step 4: Substitute the expression for 2A into the equation for B. B = (1 - (2 A)/(1 - ^2 A))1 + ((2 A)/(1 - ^2 A)) Step 5: Simplify the complex fraction by multiplying the numerator and denominator by (1 - ^2 A). B = ((1 - ^2 A) · 1 - (1 - ^2 A) · (2 A)/(1 - ^2 A))(1 - ^2 A) · 1 + (1 - ^2 A) · ((2 A)/(1 - ^2 A)) B = ((1 - ^2 A) - 2 A)/((1 - ^2 A) + 2 A) Step 6: Rearrange the terms in the numerator and denominator. B = (1 - 2 A - ^2 A)/(1 + 2 A - ^2 A) This matches the expression given by the student, thus showing how it holds. 3 done, 2 left today. You're making progress.