Question

3(a) In the diagram, BCD is an isosceles triangle in which angle BCD = 108 deg. DE is a straight line perpendicular to side DB at D. Calculate angle CDE. (b) If tan x = sqrt3 for 0 < x < 90 deg, find (cos^2 x - sin x) / (sin^2 x + cos x).

Accepted Answer

(a) Since BCD is isosceles with angle BCD=108 deg, the base angles are equal: angle CDB = angle CBD. Since the angles in a triangle sum to 180 deg: 2(angle CDB) + 108 = 180, so angle CDB = 36 deg. Since angle CDE = angle CDB + angle BDE (where angle BDE = 90 deg, as DE is perpendicular to DB), angle CDE = 36 + 90 = 126 deg.
(b) Using the Pythagoras theorem with tan x = sqrt3/1: Hypotenuse = sqrt(1^2+(sqrt3)^2) = sqrt(1+3) = 2. So cos x = 1/2, sin x = sqrt3/2. Substituting into the expression: [(1/2)^2 - (sqrt3/2)] / [(sqrt3/2)^2 + (1/2)] = [1/4 - sqrt3/2] / [3/4 + 1/2] = [(1-2sqrt3)/4] / [5/4] = (1-2sqrt3)/5.

WAEC Mathematics 2020 Theory — Question 3

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