Question

If $b=14,c=11$, and $A={60}^o$, find B and C, given that
$log2=.30103$, $log3=.4771213$,
$L\tan {11}^o{44}^{\prime }=9.3174299$,
and $L\tan {11}^o{45}^{\prime }=9.3180640$.

Answer 1

$\tan \frac{B-C}{2}=\frac{b-c}{b+c}\cot \frac{A}{2}$

$=\frac{14-11}{14+11}\cot \frac{{60}^o}{2}$

$=\frac{3}{25}\cot {30}^o$

$=\frac{3}{25}\sqrt{3}$

$=4\times \frac{3^{3/2}}{100}$

$L\tan \frac{B-C}{2}=10+2\log 2+\frac{3}{2}\log 3-\log 100$

$=10+2\times 0.30103+\frac{3}{2}\times 0.4771213-2$

$=9.3177419$

$L\tan {11}^o{45}^{\prime }-L\tan {11}^o{44}^{\prime }=9.3180640-9.3174299$

$=0.0006341$

and $L\tan \frac{B-C}{2}-L\tan {11}^o{44}^{\prime }=9.3177419-9.3174299$

$=0.0003120$

$\therefore$ Difference for $0.0003120=\frac{60\times 0.0003120}{0.0006341}=30"$

$\therefore$ $\frac{B-C}{2}={11}^o{44}^{\prime }30"$ ---- ( 1 )

And $\frac{B+C}{2}={90}^o-{30}^o$ ----- ( 2 )

Solving equations ( 1 ) and ( 2 ), we get

$\angle B={71}^o{44}^{\prime }30",$ $\angle C={48}^o{15}^{\prime }30"$