Write the names of two parallel quadrilateral and their adjacent sides and angles. Also write the relation between adjacent sides and angles.

Mathematics
Write the names of two parallel quadrilateral and their adjacent sides and angles. Also write the relation between adjacent sides and angles.

ScanSolve AI Answer

Step 1: ∠C\angle C를 구한다.

∠C=180∘−30∘−45∘=105∘\angle C = 180^\circ - 30^\circ - 45^\circ = 105^\circ

Step 2: 사인 법칙을 이용한다.

BCsin⁡30∘=ACsin⁡45∘=ABsin⁡105∘=13sin⁡105∘\frac{BC}{\sin 30^\circ} = \frac{AC}{\sin 45^\circ} = \frac{AB}{\sin 105^\circ} = \frac{13}{\sin 105^\circ} sin⁡105∘=sin⁡(60∘+45∘)=sin⁡60∘cos⁡45∘+cos⁡60∘sin⁡45∘=32⋅22+12⋅22=6+24\sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} 13sin⁡105∘=13÷6+24=13⋅46+2=526+2\frac{13}{\sin 105^\circ} = 13 \div \frac{\sqrt{6} + \sqrt{2}}{4} = 13 \cdot \frac{4}{\sqrt{6} + \sqrt{2}} = \frac{52}{\sqrt{6} + \sqrt{2}}

분모를 유리화한다:

526+2⋅6−26−2=52(6−2)6−2=52(6−2)4=13(6−2)\frac{52}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{52(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{52(\sqrt{6} - \sqrt{2})}{4} = 13(\sqrt{6} - \sqrt{2})

Step 2-(1): BCBC의 길이

sin⁡30∘=12\sin 30^\circ = \frac{1}{2} BC=13(6−2)⋅12=132(6−2) cmBC = 13(\sqrt{6} - \sqrt{2}) \cdot \frac{1}{2} = \frac{13}{2}(\sqrt{6} - \sqrt{2})\ cm

\dfrac{13{2}(\sqrt{6}-\sqrt{2})\ cm}

Step 2-(2): ACAC의 길이

sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2} AC=13(6−2)⋅22=132(6⋅2−2⋅2)=132(12−2)AC = 13(\sqrt{6} - \sqrt{2}) \cdot \frac{\sqrt{2}}{2} = \frac{13}{2} \left( \sqrt{6} \cdot \sqrt{2} - \sqrt{2} \cdot \sqrt{2} \right) = \frac{13}{2} (\sqrt{12} - 2) 12=23\sqrt{12} = 2\sqrt{3} AC=132(23−2)=13(3−1) cmAC = \frac{13}{2} (2\sqrt{3} - 2) = 13(\sqrt{3} - 1)\ cm

13(\sqrt{3-1)\ cm}

Step 3: 삼각형의 면적을 구한다.

S=12⋅AB⋅AC⋅sin⁡30∘=12⋅13⋅13(3−1)⋅12S = \frac{1}{2} \cdot AB \cdot AC \cdot \sin 30^\circ = \frac{1}{2} \cdot 13 \cdot 13(\sqrt{3} - 1) \cdot \frac{1}{2} =13⋅13(3−1)4=169(3−1)4 cm2= \frac{13 \cdot 13(\sqrt{3} - 1)}{4} = \frac{169(\sqrt{3} - 1)}{4}\ cm^2

\dfrac{169(\sqrt{3-1)}{4}\ cm^{2}}

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