1. **Problem statement:** Find the length of side JK in each triangle given angles and sides. 2. **Key formulas and rules:** - Sum of angles in a triangle is $180^\circ$. - Use Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a,b,c$ are sides opposite angles $A,B,C$ respectively. - For right triangles, trigonometric ratios (sine, cosine) can be used. --- ### a) Triangle JMN - Given: $\angle N=40^\circ$, $\angle M=20^\circ$, side $JM=5.0$ cm. - Find $JK$ (assuming $JK$ corresponds to side opposite $\angle N$ or $\angle M$; here, $JK$ is side opposite $\angle N$). **Step 1:** Find $\angle J$: $$\angle J = 180^\circ - 40^\circ - 20^\circ = 120^\circ$$ **Step 2:** Using Law of Sines: $$\frac{JK}{\sin 20^\circ} = \frac{5.0}{\sin 120^\circ}$$ **Step 3:** Solve for $JK$: $$JK = \frac{5.0 \times \sin 20^\circ}{\sin 120^\circ}$$ Calculate values: $$\sin 20^\circ \approx 0.3420, \quad \sin 120^\circ \approx 0.8660$$ $$JK = \frac{5.0 \times 0.3420}{0.8660} \approx 1.975$$ **Answer:** $JK \approx 2.0$ cm (to nearest tenth). --- ### b) Right triangle JAK - Given: $\angle J=15^\circ$, side $AK=3.0$ cm, $\angle B=60^\circ$ (assuming $B$ is $K$ or related; since right angle at $A$, $JK$ is hypotenuse). **Step 1:** Since right angle at $A$, $JK$ is hypotenuse opposite right angle. **Step 2:** Use sine of $15^\circ$: $$\sin 15^\circ = \frac{AK}{JK}$$ **Step 3:** Solve for $JK$: $$JK = \frac{AK}{\sin 15^\circ} = \frac{3.0}{\sin 15^\circ}$$ Calculate: $$\sin 15^\circ \approx 0.2588$$ $$JK = \frac{3.0}{0.2588} \approx 11.58$$ **Answer:** $JK \approx 11.6$ cm. --- ### c) Triangle JKD (right angle at C) - Given: side $KC=3.0$ cm, $\angle K=35^\circ$, $\angle D=30^\circ$. - Find $JK$. **Step 1:** Find $\angle J$: $$\angle J = 180^\circ - 90^\circ - 35^\circ = 55^\circ$$ **Step 2:** Use Law of Sines: $$\frac{JK}{\sin 30^\circ} = \frac{KC}{\sin 55^\circ}$$ **Step 3:** Solve for $JK$: $$JK = \frac{KC \times \sin 30^\circ}{\sin 55^\circ} = \frac{3.0 \times 0.5}{0.8192} \approx 1.83$$ **Answer:** $JK \approx 1.8$ cm. --- ### d) Triangle FJE - Given: $\angle FJ=60^\circ$, $\angle KJ=35^\circ$, side $FE=4.2$ cm. - Find $JK$. **Step 1:** Find third angle: $$\angle J = 180^\circ - 60^\circ - 35^\circ = 85^\circ$$ **Step 2:** Use Law of Sines: $$\frac{JK}{\sin 60^\circ} = \frac{FE}{\sin 85^\circ}$$ **Step 3:** Solve for $JK$: $$JK = \frac{4.2 \times \sin 60^\circ}{\sin 85^\circ}$$ Calculate: $$\sin 60^\circ \approx 0.8660, \quad \sin 85^\circ \approx 0.9962$$ $$JK = \frac{4.2 \times 0.8660}{0.9962} \approx 3.65$$ **Answer:** $JK \approx 3.7$ cm. --- **Final answers:** - a) $2.0$ cm - b) $11.6$ cm - c) $1.8$ cm - d) $3.7$ cm