Question

1. a) in \\( \triangle a b c, a b = 18.7 \mathrm { cm }, a c = 17.9 \mathrm { cm } \\), and \\( \angle b = 70 ^ { \circ } \\);

determine the measure of bc to the nearest tenth of a
centimetre.
\\( \frac { c } { \sin c } = \frac { b } { \sin b } \\)
\\( \frac { 18.7 } { \sin c } = \frac { 17.9 } { \sin 70 } \\)
\\( \angle c = \sin ^ { - 1 } \left( \frac { 18.7 \sin 70 } { 17.9 } \
ight) \\)
\\( \angle c = 79 ^ { \circ } \\)
\\( \angle a = 79 ^ { \circ } - 70 ^ { \circ } \\)
\\( \angle a = 149 ^ { \circ } \\)
\\( \angle a = 180 ^ { \circ } - 149 ^ { \circ } \\)
\\( \angle a = 31 ^ { \circ } \\)
\\( \frac { a } { \sin a } = \frac { b } { \sin b } \\)
\\( \frac { a } { \sin 31 ^ { \circ } } = \frac { 17.9 } { \sin 70 } \\)
\\( a = \frac { 17.9 \sin 31 } { \sin 70 } \\)
\\( a = 9.8 \mathrm { cm } \\)

Explanation:

Step1: Apply Law of Sines for angle C

$\frac{\sin C}{AB} = \frac{\sin B}{AC} \implies \sin C = \frac{AB \cdot \sin B}{AC} = \frac{18.7 \cdot \sin 70^\circ}{17.9}$
$\sin C \approx \frac{18.7 \cdot 0.9397}{17.9} \approx \frac{17.572}{17.9} \approx 0.9817$
$\angle C \approx \sin^{-1}(0.9817) \approx 79^\circ$ (valid, since $79^\circ < 180^\circ - 70^\circ = 110^\circ$)

Step2: Calculate angle A

$\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 70^\circ - 79^\circ = 31^\circ$

Step3: Apply Law of Sines for BC

$\frac{BC}{\sin A} = \frac{AC}{\sin B} \implies BC = \frac{AC \cdot \sin A}{\sin B} = \frac{17.9 \cdot \sin 31^\circ}{\sin 70^\circ}$
$\sin 31^\circ \approx 0.5150, \sin 70^\circ \approx 0.9397$
$BC \approx \frac{17.9 \cdot 0.5150}{0.9397} \approx \frac{9.2185}{0.9397} \approx 9.8$

Answer:

9.8 cm