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For part (i), my answer is correct but my answer for (ii) seems to be a little bit out. I can't spot where I've gone wrong. Can anyone help me out?
Many thanks.
Q. In the given triangle, find (i) [itex]\angle abc[/itex], (ii) [itex]\angle bac[/itex].
(i) cos B = [itex]\frac{c^2+a^2b^2}{2ac}[/itex] => cos B = [itex]\frac{5^2+8^26^2}{(2)(8)(5)}[/itex] => cos B = [itex]\frac{53}{80}[/itex] => B = secant 0.6625 => B = [itex]48^o 31'[/itex]
(ii) [itex]\frac{sin48^o 31'}{6} = \frac{sin x}{8}[/itex] => [itex]\frac{0.7491}{6} = \frac{sin x}{8}[/itex] => sin x = (8)(0.1249) => x = cosec 0.9989 => x = [itex]87^o42'[/itex]
Ans.: (From text book): (i) [itex]48^o 31'[/itex], (ii) [itex]92^o 52'[/itex]
Many thanks.
Homework Statement
Q. In the given triangle, find (i) [itex]\angle abc[/itex], (ii) [itex]\angle bac[/itex].
The Attempt at a Solution
(i) cos B = [itex]\frac{c^2+a^2b^2}{2ac}[/itex] => cos B = [itex]\frac{5^2+8^26^2}{(2)(8)(5)}[/itex] => cos B = [itex]\frac{53}{80}[/itex] => B = secant 0.6625 => B = [itex]48^o 31'[/itex]
(ii) [itex]\frac{sin48^o 31'}{6} = \frac{sin x}{8}[/itex] => [itex]\frac{0.7491}{6} = \frac{sin x}{8}[/itex] => sin x = (8)(0.1249) => x = cosec 0.9989 => x = [itex]87^o42'[/itex]
Ans.: (From text book): (i) [itex]48^o 31'[/itex], (ii) [itex]92^o 52'[/itex]
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