Question

using the law of sines for the ambiguous case
$\triangle jkl$ has $j=7$, $k=11$, and $mangle j=18^{circ}$. complete the statements to determine all possible measures of angle k
triangle jkl meets the $\boldsymbol{square}$ criteria, which means it is the ambiguous case.
substitute the known values into the law of sines: $\frac{sin(18^{circ})}{7} = \frac{sin(k)}{11}$.
cross multiply $11\sin(18^{circ}) = \boldsymbol{square}$.
solve for the measure of angle k, and use a calculator to determine the value.
round to the nearest degree: $mangle k approx \boldsymbol{square}^{circ}$
however, because this is the ambiguous case, the measure of angle k could also be $\boldsymbol{square}^{circ}$

Explanation:

Step1: Identify ambiguous case criteria

The ambiguous case (SSA) occurs when we have two sides and a non-included angle, where the shorter side is opposite the given angle. Here, $j=7$ (opposite $\angle J=18^\circ$) is shorter than $k=11$.

Step2: Cross-multiply Law of Sines

Starting with $\frac{\sin(18^\circ)}{7} = \frac{\sin(K)}{11}$, cross-multiply to isolate the term with $\sin(K)$.
$11\sin(18^\circ) = 7\sin(K)$

Step3: Solve for $\sin(K)$

Rearrange to solve for $\sin(K)$:
$\sin(K) = \frac{11\sin(18^\circ)}{7}$
Calculate $\sin(18^\circ) \approx 0.3090$, so:
$\sin(K) \approx \frac{11 \times 0.3090}{7} \approx \frac{3.399}{7} \approx 0.4856$
Find $\angle K$ using inverse sine:
$\angle K \approx \sin^{-1}(0.4856) \approx 29^\circ$

Step4: Find second possible angle

In the ambiguous case, a second angle is $180^\circ -$ the acute angle found.
$180^\circ - 29^\circ = 151^\circ$
Verify the sum with $\angle J$: $151^\circ + 18^\circ = 169^\circ < 180^\circ$, so it is valid.

Answer:

Triangle JKL meets the SSA (side-side-angle, shorter side opposite given angle) criteria, which means it is the ambiguous case.
Cross multiply $11\sin(18^{\circ}) = \boldsymbol{7\sin(K)}$
Round to the nearest degree: $m\angle K \approx \boldsymbol{29}^{\circ}$
However, because this is the ambiguous case, the measure of angle K could also be $\boldsymbol{151}^{\circ}$