WASSCE Further Maths: Geometry & Trigonometry Practice Questions & Answers

The compound angle formula for the sine of a difference is a standard trigonometric identity: $\sin(A - B) = \sin A \cos B - \cos A \sin B$sin(A−B)=sinAcosB−cosAsinB.

Parallel lines have the same gradient. Rearranging $3x - 2y + 5 = 0$3x−2y+5=0 gives $y = \frac{3}{2}x + \frac{5}{2}$y=23​x+25​, so the gradient $m = \frac{3}{2}$m=23​. Using the point-slope form: $y - (-2) = \frac{3}{2}(x - 1) \implies 2(y + 2) = 3(x - 1) \implies 2y + 4 = 3x - 3 \implies 3x - 2y - 7 = 0$y−(−2)=23​(x−1)⟹2(y+2)=3(x−1)⟹2y+4=3x−3⟹3x−2y−7=0.

Differentiating $y^2 = 4x$y2=4x implicitly yields $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$2ydxdy​=4⟹dxdy​=y2​. At $(1, 2)$(1,2), the tangent gradient is $m = \frac{2}{2} = 1$m=22​=1. The normal is perpendicular to the tangent, so its gradient $m' = -1$m′=−1. The equation is $y - 2 = -1(x - 1) \implies y + x = 3$y−2=−1(x−1)⟹y+x=3.

Use the identity $\cos^2\theta = 1 - \sin^2\theta$cos2θ=1−sin2θ. Substitute to get $2(1 - \sin^2\theta) + \sin\theta - 1 = 0 \implies 2 - 2\sin^2\theta + \sin\theta - 1 = 0 \implies 2\sin^2\theta - \sin\theta - 1 = 0$2(1−sin2θ)+sinθ−1=0⟹2−2sin2θ+sinθ−1=0⟹2sin2θ−sinθ−1=0. Factoring gives $(2\sin\theta + 1)(\sin\theta - 1) = 0$(2sinθ+1)(sinθ−1)=0. So, $\sin\theta = -0.5$sinθ=−0.5 (no solution in $0^\circ \le \theta \le 180^\circ$0∘≤θ≤180∘) or $\sin\theta = 1 \implies \theta = 90^\circ$sinθ=1⟹θ=90∘.

The midpoint formula is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$(2x1​+x2​​,2y1​+y2​​). Thus, $\left(\frac{-4 + 2}{2}, \frac{6 + (-2)}{2}\right) = \left(\frac{-2}{2}, \frac{4}{2}\right) = (-1, 2)$(2−4+2​,26+(−2)​)=(2−2​,24​)=(−1,2).

The standard equation of a parabola opening to the right is $y^2 = 4ax$y2=4ax, where the focus is at $(a, 0)$(a,0). Comparing this with $y^2 = 16x$y2=16x, we get $4a = 16 \implies a = 4$4a=16⟹a=4. Therefore, the focus is $(4, 0)$(4,0).

Express $75^\circ$75∘ as $45^\circ + 30^\circ$45∘+30∘. Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$cos(A+B)=cosAcosB−sinAsinB, we have $\cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}$cos(45∘+30∘)=cos45∘cos30∘−sin45∘sin30∘=(22​​)(23​​)−(22​​)(21​)=46​−2​​.

Let the x-axis divide the segment in the ratio $k:1$k:1. Any point on the x-axis has a y-coordinate of $0$0. Using the section formula for the y-coordinate: $y = \frac{k y_2 + 1 y_1}{k + 1} = 0 \implies \frac{k(6) + 1(-3)}{k + 1} = 0 \implies 6k - 3 = 0 \implies k = \frac{1}{2}$y=k+1ky2​+1y1​​=0⟹k+1k(6)+1(−3)​=0⟹6k−3=0⟹k=21​. The ratio is $\frac{1}{2}:1$21​:1, which simplifies to $1:2$1:2.

The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$(x−h)2+(y−k)2=r2, where the center is $(h, k)$(h,k). Comparing the given equation to the standard form, $h = 5$h=5 and $k = -3$k=−3. Thus, the center is $(5, -3)$(5,−3).

Use the identity $\tan^2 x = \sec^2 x - 1$tan2x=sec2x−1. Substitute into the equation: $\sec^2 x - 1 + \sec x = 1 \implies \sec^2 x + \sec x - 2 = 0$sec2x−1+secx=1⟹sec2x+secx−2=0. Factoring this quadratic in $\sec x$secx yields $(\sec x + 2)(\sec x - 1) = 0$(secx+2)(secx−1)=0. Thus, $\sec x = -2 \implies \cos x = -\frac{1}{2}$secx=−2⟹cosx=−21​ (which gives $x = 120^\circ, 240^\circ$x=120∘,240∘) or $\sec x = 1 \implies \cos x = 1$secx=1⟹cosx=1 (which gives $x = 0^\circ$x=0∘).