Find values for b such that the triangle has two solutions. A = 32°, a = 5 A) $5 \leq b \leq \frac { 5 } { \sin 32 ^ { \circ } }$ B) $5

Question

Find values for b such that the triangle has two solutions. A = 32°, a = 5 A) $5 \leq b \leq \frac { 5 } { \sin 32 ^ { \circ } }$ B) $5 < b < \frac { 5 } { \sin 32 ^ { \circ } }$ C) $5 < b \leq \frac { 5 } { \sin 32 ^ { \circ } }$ D) $5 \geq b \geq \frac { 5 } { \sin 32 ^ { \circ } }$ E) $5 > b > \frac { 5 } { \sin 32 ^ { \circ } }$

Quizplus · Accepted Answer

For a triangle to have two solutions in the context of the Law of Sines, the side opposite the given angle (in this case, side a with angle A = 32°) must be shorter than the side length b, but longer than the altitude from the angle opposite b. This means $a < b$ and $a > b \sin A$, which simplifies to $5 < b$ and $5 > b \sin 32^\circ$. Rearranging the second inequality gives $5 < \frac{5}{\sin 32^\circ}$. Therefore, the correct range for b is $5 < b < \frac{5}{\sin 32^\circ}$.

Find Values for B Such That the Triangle Has Two 5≤b≤5sin⁡32∘5 \leq b \leq \frac { 5 } { \sin 32 ^ { \circ } }5≤b≤sin32∘5​

Find Values for B Such That the Triangle Has Two $5 \leq b \leq \frac { 5 } { \sin 32 ^ { \circ } }$