(a) Given that \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). Without using mathematical tables or calculator, evaluate \(\sin 105°\), leaving your answer...

Assessment: WAEC SSCE - General Mathematics - 1999 (Essay) Subject: General Mathematics

Question 1 Report

(a) Given that \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). Without using mathematical tables or calculator, evaluate \(\sin 105°\), leaving your answer in the surd form.

(You may use 105° = 60° + 45°)

(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?

(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \(2\) and \(\frac{2}{9}\) respectively. Find :

(i) the common difference ; (ii) the 5th term.

Answer Details

(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]

(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.

(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]

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