Question

12.c solve the following problems. given: △klm, lm = 12, m∠k = 60°, m∠m = 45° find: the perimeter of △klm answer: p_{klm}=

Explanation:

Step1: Find angle L

The sum of angles in a triangle is 180°. So, $m\angle L=180^{\circ}-m\angle K - m\angle M=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$

Step2: Use the Law of Sines to find KL

By the Law of Sines, $\frac{KL}{\sin M}=\frac{LM}{\sin K}$. Substituting the given values: $\frac{KL}{\sin45^{\circ}}=\frac{12}{\sin60^{\circ}}$. Then $KL = \frac{12\sin45^{\circ}}{\sin60^{\circ}}=\frac{12\times\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}=\frac{12\sqrt{2}}{\sqrt{3}} = 4\sqrt{6}$

Step3: Use the Law of Sines to find KM

By the Law of Sines, $\frac{KM}{\sin L}=\frac{LM}{\sin K}$. First, $\sin75^{\circ}=\sin(45^{\circ} + 30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$. Then $\frac{KM}{\sin75^{\circ}}=\frac{12}{\sin60^{\circ}}$, so $KM=\frac{12\sin75^{\circ}}{\sin60^{\circ}}=\frac{12\times\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{3}}{2}} = 6 + 2\sqrt{3}$

Step4: Calculate the perimeter

The perimeter $P = KL+LM + KM=4\sqrt{6}+12+(6 + 2\sqrt{3})=18 + 2\sqrt{3}+4\sqrt{6}$

Answer:

$18 + 2\sqrt{3}+4\sqrt{6}$