Question

12.e solve the following problems. given: △mno, m∠m = 45°, m∠o = 30°, mn = 6 find: no, mo answer: no = , mo = clear all

Explanation:

Step1: Find angle N

The sum of angles in a triangle is 180°. So, $m\angle N=180^{\circ}-(45^{\circ} + 30^{\circ})=105^{\circ}$.

Step2: Use the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. For $\triangle MNO$, we have $\frac{NO}{\sin M}=\frac{MN}{\sin O}=\frac{MO}{\sin N}$.

Step3: Find NO

We know that $\frac{NO}{\sin M}=\frac{MN}{\sin O}$. Substituting $M = 45^{\circ}$, $O=30^{\circ}$ and $MN = 6$, we get $\frac{NO}{\sin45^{\circ}}=\frac{6}{\sin30^{\circ}}$. Then $NO=\frac{6\sin45^{\circ}}{\sin30^{\circ}}$. Since $\sin45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin30^{\circ}=\frac{1}{2}$, $NO = 6\sqrt{2}$.

Step4: Find MO

We know that $\frac{MO}{\sin N}=\frac{MN}{\sin O}$. First, $\sin105^{\circ}=\sin(60^{\circ}+ 45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}+ 1}{2\sqrt{2}}$. Then $\frac{MO}{\sin105^{\circ}}=\frac{6}{\sin30^{\circ}}$, so $MO=\frac{6\sin105^{\circ}}{\sin30^{\circ}}$. Substituting $\sin105^{\circ}=\frac{\sqrt{3}+ 1}{2\sqrt{2}}$ and $\sin30^{\circ}=\frac{1}{2}$, we get $MO = 3(\sqrt{3}+ 1)$.

Answer:

$NO = 6\sqrt{2}$, $MO=3(\sqrt{3}+ 1)$