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For part

Many thanks.

**(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.

## 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^2-b^2}{2ac}[/itex] => cos B = [itex]\frac{5^2+8^2-6^2}{(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]