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) must be shorter than the side b, but longer than the altitude from the angle opposite side b. This means $5 < b < \frac{5}{\sin 32^\circ}$, where $b$ cannot equal $5$ or $\frac{5}{\sin 32^\circ}$ to ensure two distinct solutions.

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 $5 \leq b \leq \frac { 5 } { \sin 32 ^ { \circ } }$