(a} In the diagram, O is the centre of the circle... | WAEC SSCE - General Mathematics - 2020 (Essay)

Question 1 Report

(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE.

(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate

\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)

Answer

(a) Finding \(\widehat{CDE}\). From the diagram \(A,B,C,D,E\) lie on the circle with centre \(O\), and the lines \(AD\) and \(EB\) both pass through \(O\), so \(AD\) and \(EB\) are diameters. Also \(|BC|=|CD|\) and \(\widehat{BCD}=108^\circ\).

Equal chords cut equal arcs, so \(\text{arc }BC=\text{arc }CD\); call each \(a\).

\(\widehat{BCD}=108^\circ\) is the angle at the circumference standing on the arc \(BAED\) (the arc from \(B\) to \(D\) not through \(C\)):

\[\text{arc }BAED=2\times108^\circ=216^\circ\;\Rightarrow\;\text{arc }BCD=360^\circ-216^\circ=144^\circ.\]

So \(2a=144^\circ\Rightarrow a=72^\circ\); thus \(\text{arc }BC=\text{arc }CD=72^\circ.\)

Since \(AD\) is a diameter, arc \(ABCD\) (semicircle) \(=180^\circ\), so \(\text{arc }AB=180^\circ-(72^\circ+72^\circ)=36^\circ.\) Since \(EB\) is a diameter, arc \(EAB=180^\circ\), so \(\text{arc }AE=180^\circ-36^\circ=144^\circ,\) and \(\text{arc }ED=180^\circ-144^\circ=36^\circ.\)

\(\widehat{CDE}\) stands on the arc \(CBAE\) (from \(C\) to \(E\) not through \(D\)):

\[\text{arc }CBAE=72^\circ+36^\circ+144^\circ=252^\circ,\qquad \widehat{CDE}=\tfrac12(252^\circ)=\boxed{126^\circ.}\]

(Check: \(\widehat{CDE}=\widehat{CDA}+\widehat{ADE}=\tfrac12(108^\circ)+\tfrac12(144^\circ)=54^\circ+72^\circ=126^\circ.\))

(b) Evaluate \(\dfrac{\cos^2 x-\sin x}{\sin^2 x+\cos x}\) given \(\tan x=\sqrt3,\ 0^\circ\le x\le 90^\circ.\)

\(\tan x=\sqrt3\Rightarrow x=60^\circ,\) so \(\sin x=\dfrac{\sqrt3}{2},\ \cos x=\dfrac12.\)

\[\text{Numerator}=\left(\tfrac12\right)^2-\tfrac{\sqrt3}{2}=\tfrac14-\tfrac{\sqrt3}{2}=\frac{1-2\sqrt3}{4}.\]\[\text{Denominator}=\left(\tfrac{\sqrt3}{2}\right)^2+\tfrac12=\tfrac34+\tfrac12=\tfrac54.\]\[\frac{\frac{1-2\sqrt3}{4}}{\frac54}=\frac{1-2\sqrt3}{5}\approx-0.49.\]

Read lesson note on Trigonometry (JAMB)

Read lesson note on Circles (WAEC)

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