Finding Triangle Angles with Cosine Rule

[odolwa99]

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.

Homework Statement

Q. In the given triangle, find (i) $|\angle abc|$, (ii) $|\angle bac|$.

The Attempt at a Solution

(i) cos B = $\frac{c^2+a^2-b^2}{2ac}$ => cos B = $\frac{5^2+8^2-6^2}{(2)(8)(5)}$ => cos B = $\frac{53}{80}$ => B = secant 0.6625 => B = $48^o 31'$

(ii) $\frac{sin48^o 31'}{6} = \frac{sin x}{8}$ => $\frac{0.7491}{6} = \frac{sin x}{8}$ => sin x = (8)(0.1249) => x = cosec 0.9989 => x = $87^o42'$

Ans.: (From textbook): (i) $48^o 31'$, (ii) $92^o 52'$

 

[Villyer]

I'm confused with cos B = 5380 => B = secant 0.6625
I think you meant arccos? secant is the inverse of cosine, so secant = 1 / cosine
arccos is the inverse function of cosine, so arccos(cos(x)) = xAlso, the answer given is larger than 90 degrees. The range of arcsin is -90 to 90, therefore you have to consider all possible results to include other possible angles.

 

[odolwa99]

So arcsin, with range $-90^o$ to $90^o = 180^o$. Thus, I noticed that $180^o-87^o42'=92^o18'$, which takes me much closer to the intended answer. But am I solving this correctly...?

 

[Villyer]

I don't have a calculator on me, but that small difference might just be a rounding error. Did you use an exact value when you took arcsin, or the rounded 0.9989?

 

[odolwa99]

Ok, I've gone back to get the exact answer and $\frac{sin48^o31'}{6}=\frac{sinx}{8}$ becomes $x=87^o16'$. So, $180^0-87^o16'=92^o44'$. Thats still a little bit out...is that acceptable?
Also, one other question, what indicator is there to know that I'm meant to subtract $180^0-87^o16'$ and not just use $87^o16'$

 

[ehild]

Apply cosine law again to the angle opposite to the side of length 6.

ehild