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

Mathematics

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Step-by-step solution

5 steps

  1. ∠C\angle C를 구한다.

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

  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})

  3. (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})\ \mathrm{cm}

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

  4. (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)\ \mathrm{cm}

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

  5. 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}\ \mathrm{cm}^2

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

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