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If $\sin(A + 2B) = 1$ and $\cos(2A + B) = \frac{1}{2}$, find the values of A and B.
Hence, find the value of $\tan (B – A)$.
Hence, find the value of $\tan (B – A)$.
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(b) $\sin (A + 2B) = 1 \Rightarrow A + 2B = 90^\circ$ --------(i) (1/2 Mark)
$\cos (2A + B) = \frac{1}{2} \Rightarrow 2A + B = 60^\circ$ --------(ii) (1/2 Mark)
Solving (i) and (ii) we get $A = 10^\circ$ and $B = 40^\circ$ (1 Mark)
$\tan (B-A) = \tan (40^\circ – 10^\circ) = \tan 30^\circ = \frac{1}{\sqrt{3}}$ (1 Mark)
(b) $\sin (A + 2B) = 1 \Rightarrow A + 2B = 90^\circ$ --------(i) (1/2 Mark)
$\cos (2A + B) = \frac{1}{2} \Rightarrow 2A + B = 60^\circ$ --------(ii) (1/2 Mark)
Solving (i) and (ii) we get $A = 10^\circ$ and $B = 40^\circ$ (1 Mark)
$\tan (B-A) = \tan (40^\circ – 10^\circ) = \tan 30^\circ = \frac{1}{\sqrt{3}}$ (1 Mark)