Find values for b such that the triangle has one solution.
A = 24 °, a = 16.2

A) $b 16.2 , b \frac { 16.2 } { \sin 24 ^ { \circ } }$
B) $b \leq 16.2 , b = \frac { 16.2 } { \sin 24 ^ { \circ } }$
C) $b \leq 16.2 , b \neq \frac { 16.2 } { \sin 24 ^ { \circ } }$
D) $b \geq 16.2 , b \geq \frac { 16.2 } { \sin 24 ^ { \circ } }$
E) $b \geq 16.2 , b = \frac { 16.2 } { \sin 24 ^ { \circ } }$

Question

Find values for b such that the triangle has one solution.  
A = 24 °, a = 16.2
 
A) $b > 16.2 , b > \frac { 16.2 } { \sin 24 ^ { \circ } }$
B) $b \leq 16.2 , b = \frac { 16.2 } { \sin 24 ^ { \circ } }$
C) $b \leq 16.2 , b \neq \frac { 16.2 } { \sin 24 ^ { \circ } }$
D) $b \geq 16.2 , b \geq \frac { 16.2 } { \sin 24 ^ { \circ } }$
E) $b \geq 16.2 , b = \frac { 16.2 } { \sin 24 ^ { \circ } }$

Quizplus · Accepted Answer

For a triangle to have one solution in the SSA case, b must be equal to the height, which is $ \frac{a}{\sin A} $.

Find Values for B Such That the Triangle Has One b>16.2,b>16.2sin⁡24∘b > 16.2 , b > \frac { 16.2 } { \sin 24 ^ { \circ } }b>16.2,b>sin24∘16.2​

Find Values for B Such That the Triangle Has One $b > 16.2 , b > \frac { 16.2 } { \sin 24 ^ { \circ } }$