Find values for b such that the triangle has two solutions. ?
A = 10°,a = 18.6
?

Question

Find values for b such that the triangle has two solutions. ? A = 10°,a = 18.6 ? A) $18.6 < b \leq \frac { 18.6 } { \sin 10 ^ { \circ } }$ B) $18.6 < b < \frac { 18.6 } { \sin 10 ^ { \circ } }$ C) $18.6 > b > \frac { 18.6 } { \sin 10 ^ { \circ } }$ D) $18.6 \leq b \leq \frac { 18.6 } { \sin 10 ^ { \circ } }$ E) $18.6 \geq b \geq \frac { 18.6 } { \sin 10 ^ { \circ } }$

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 hypotenuse (the longest side in a right triangle, but in this context, it's the side opposite the largest angle) but longer than the projection of side a onto the side opposite the given angle A. This means b must be greater than a but less than a divided by the sine of angle A. Therefore, the correct condition is $18.6 < b < \frac { 18.6 } { \sin 10 ^ { \circ } }$.

Find Values for B Such That the Triangle Has Two 18.6<b≤18.6sin⁡10∘18.6 < b \leq \frac { 18.6 } { \sin 10 ^ { \circ } }18.6<b≤sin10∘18.6​

Find Values for B Such That the Triangle Has Two $18.6 < b \leq \frac { 18.6 } { \sin 10 ^ { \circ } }$