1. Solving Trig Equations Example (i): Step 1: Expand sin(a-b) Equate coefficients Square & add eqns 1&2 Subst. for k in eqn 2 c a s t Take 2 nd quadrant:

2. 2x-146.3 = …., 16.1 o, 180-16.1 o, 360+16.1 o, 360+180-16.1 o, …. 2x-146.3 = …., 16.1 o, 163.9 o, 376.1 o, 523.9 o, …. 2x = …., 162.4 o, 310.2 o, 522.4 o, 670.2 o, …. c a s t x = …., 81.2 o, 155.1 o, 261.2 o, 335.1 o, …. Return to Main

3. Example (ii): (From a past paper) A builder has obtained a large supply of 4 metre rafters. He wishes to use them to build some holiday chalets. The planning department insists that the gable end of each chalet should be in the form of an isosceles triangle surmounting two squares, as shown in the diagram. a) If  o is the angle shown in the diagram and A is the area (in square metres) of the gable end, show that c) Find algebraically the value of  for which the area of the gable end is 30 square metres. 4 4

4. 4 4 s s Let the side of the square frames be s. Part (a) Use the cosine rule in the isosceles triangle: This is the area of one of the squares. Use the formula for the area of an isosceles triangle. Total area = Triangle + 2 x square:

5. Part (b) Expand sin(a+b) Equate coefficients. Square & add eqns 1 & 2 Only interested in +ve root. Subst. for t in eqn 1. Only interested in 1 st quadrant. Finally:

6. Part (c) Find algebraically the value of  o for which the area is the 30m 2 Return to Main