Perbandingan Trigonometri

Panjang sisi AC dapat kita cari menggunakan rumus pythagoras \(\text{AB}^2 + \text{BC}^2 = \text{AC}^2\) \(3^2 + 4^2 = \text{AC}^2\) \(9 + 16 = \text{AC}^2\) \(25 = \text{AC}^2\) \(\text{AC} = \sqrt{25} = 5\)

Panjang sisi BC dapat kita cari menggunakan rumus pythagoras \(\text{AB}^2 + \text{BC}^2 = \text{AC}^2\) \((2\sqrt{2})^2 + \text{BC}^2 = (2\sqrt{5})^2\) \(8 + \text{BC}^2 = 20\) \(\text{BC}^2 = 20-8\) \(\text{BC}^2 = 12\) \(\text{BC} = \sqrt{12}\) \(\text{BC} = \sqrt{4\times 3}\) \(\text{BC} = 2\sqrt{3}\)

Panjang sisi BC dapat kita cari menggunakan rumus pythagoras \(\text{AB}^2 + \text{BC}^2 = \text{AC}^2\) \(1^2 + \text{BC}^2 = p^2\) \(1 + \text{BC}^2 = p^2\) \(\text{BC}^2 = p^2-1\) \(\text{BC} = \sqrt{p^2-1}\)

Mencari panjang AB Lihat segitiga siku-siku ABD \(\tan 45^{\circ} = \dfrac{\text{de}}{\text{sa}}=\dfrac{\text{AB}}{\text{BD}}\) \(1 = \dfrac{\text{AB}}{8}\) \(\text{AB} = 8 \text{ cm}\) Mencari panjang AC Lihat segitiga siku-siku ABC \(\sin 30^{\circ} = \dfrac{\text{de}}{\text{mi}} = \dfrac{\text{AB}}{\text{AC}}\) \(\dfrac{1}{2} =  \dfrac{8}{\text{AC}}\) \(\text{AC} = 16 \text{ cm}\) Mencari panjang DC Lihat segitiga siku-siku ABC, kita dapat mencari panjang BC terlebih dahulu \(\tan 30^{\circ} = \dfrac{\text{de}}{\text{sa}} = \dfrac{\text{AB}}{\text{BC}}\) \(\dfrac{\sqrt{3}}{3} = \dfrac{8}{\text{BC}}\) \(\sqrt{3}\cdot \text{BC} = 24\) \(\text{BC} = \dfrac{24}{\sqrt{3}} \times \color{blue} \dfrac{\sqrt{3}}{\sqrt{3}}\) \(\text{BC} = \dfrac{24\sqrt{3}}{3}\) \(\text{BC} = 8\sqrt{3}\) \(\text{DC = BC – BD}\) \(\text{DC} = 8\sqrt{3} – 8 \text{ cm}\)

Mencari Panjang DB Perhatikan segitiga siku-siku DBC \(\tan 60^{\circ} = \dfrac{\text{BC}}{\text{DB}}\) \(\sqrt{3} = \dfrac{\text{BC}}{\text{DB}}\) \(\text{BC} = \sqrt{3}\cdot\text{DB}\dotso\dotso(1)\) Perhatikan segitiga siku-siku ABC \(\tan 30^{\circ} = \dfrac{\text{BC}}{\text{AB}}\:\:\:\:\:\color{blue} \text{substitusikan BC dari persamaan 1}\) \(\dfrac{\sqrt{3}}{3} = \dfrac{ \sqrt{3}\cdot\text{DB}}{10 + \text{DB}}\) \(\dfrac{\cancel{\sqrt{3}}}{3} = \dfrac{\cancel{ \sqrt{3}}\cdot\text{DB}}{10 + \text{DB}}\) \(\dfrac{1}{3} = \dfrac{\text{DB}}{10 + \text{DB}}\:\:\:\:\:\color{blue}\text{kali silang}\) \(10 + \text{DB} = 3\text{DB}\) \(10 = 3\text{DB} – \text{DB} \) \(10 = 2\text{DB}\) \(\text{DB} = \dfrac{10}{2} = 5 \text{ cm}\) Mencari Panjang BC \(\text{Dari persamaan 1:}\) \(\text{BC} = \sqrt{3}\cdot\text{DB}\) \(\text{BC} = \sqrt{3}\times 5\) \(\text{BC} = 5\sqrt{3}\text{ cm}\) Mencari Panjang AC Lihat segitiga siku-siku ABC \(\sin 30^{\circ} = \dfrac{\text{BC}}{\text{AC}}\) \(\dfrac{1}{2} = \dfrac{ 5\sqrt{3}}{\text{AC}}\:\:\:\:\:\color{blue}\text{kali silang}\) \(\text{AC} = 2\times 5\sqrt{3}\) \(\text{AC} = 10\sqrt{3}\text{ cm}\)

Mencari Panjang BE Lihat segitiga siku-siku BEC \(\sin 45^{\circ} = \dfrac{\text{BE}}{\text{BC}}\) \(\dfrac{\sqrt{2}}{2} = \dfrac{\text{BE}}{10\sqrt{2}}\:\:\:\:\:\color{blue}\text{kalikan silang}\) \(\sqrt{2}\times 10\sqrt{2} = 2 \times \text{BE}\) \(\cancel{2}\times 10 = \cancel{2} \times \text{BE}\) \(\text{BE} = 10 \text{ cm}\) Panjang \(\text{EC = BE}\) karena \(\triangle \text{BEC}\) adalah segitiga sama kaki Mencari Panjang AB Lihat segitiga siku-siku AEB \(\sin 60^{\circ} = \dfrac{\text{BE}}{\text{AB}}\) \(\dfrac{\sqrt{3}}{2}= \dfrac{10}{\text{AB}}\:\:\:\:\:\color{blue}\text{kalikan silang}\) \(\sqrt{3}\cdot\text{AB} = 20\) \(\text{AB} = \dfrac{20}{\sqrt{3}}\) \(\text{AB} = \dfrac{20}{\sqrt{3}}\times \color{blue} \dfrac{\sqrt{3}}{\sqrt{3}}\) \(\text{AB} = \dfrac{20}{3}\sqrt{3}\text{ cm}\) Mencari Panjang AC Lihat segitiga siku-siku AEB Kita cari dahulu panjang AE, \(\cos 60^{\circ} = \dfrac{\text{AE}}{\text{AB}}\) \(\dfrac{1}{2} = \dfrac{\text{AE}}{\dfrac{20}{3}\sqrt{3}}\) \(\dfrac{1}{2} = \dfrac{3\text{AE}}{20\sqrt{3}}\:\:\:\:\:\color{blue}\text{kalikan silang}\) \(\cancelto{10}{20}\sqrt{3} = \cancel{2} \times 3\text{AE}\) \(10\sqrt{3}  = 3\text{AE}\) \(\text{AE} = \dfrac{10\sqrt{3} }{3} \text{ cm}\) \(\text{AC} = \text{AE} + \text{EC}\) \(\text{AC} = \dfrac{10\sqrt{3} }{3} +10 \text{ cm}\)

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Jarak mata Galvin diukur dari permukaan tanah adalah \(\text{DC} = 150 \text{ cm} = 1,5 \text{ m}\) Jarak antara Galvin dengan tiang bendera adalah \(\text{BC} = 11\sqrt{3}\) m Tinggi tiang bendera adalah \(\text{AB}\) Untuk mencari tinggi tiang bendera, perhatikan \(\triangle \text{AED}\) \(\tan 30^{\circ} = \dfrac{\text{AE}}{\text{ED}}\) \(\dfrac{1}{\sqrt{3}} = \dfrac{\text{AE}}{11\sqrt{3}}\:\:\:\:\:\color{blue}\text{kalikan silang}\) \(11\cancel{\sqrt{3}} = \cancel{\sqrt{3}}\cdot \text{AE}\) \(\text{AE} = 11 \text{ m}\) \(\text{AB} = \text{AE} + \text{EB}\) \(\text{AB} = 11 + 1,5\) \(\text{AB} = 12,5 \text{ m}\) Jadi tinggi tiang bendera tersebut adalah \(12,5 \text{ m}\)

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Langkah mencari tinggi tower \(\text{AB}\) Perhatikan \(\triangle \text{ABD}\) \(\tan 60^{\circ} = \dfrac{\text{AB}}{\text{BD}}\) \(\sqrt{3} = \dfrac{\text{AB}}{\text{BD}}\) \(\text{AB} = \sqrt{3}\cdot \text{BD}\) Perhatikan \(\triangle \text{ABC}\) \(\tan 45^{\circ} = \dfrac{\text{AB}}{\text{BC}}\) \(1 = \dfrac{\text{AB}}{\text{BC}}\) \(\text{BC} = \text{AB}\) \(\text{BD} + 20 = \sqrt{3}\cdot\text{BD}\) \(20 = \sqrt{3}\cdot\text{BD} – \text{BD}\) \(20 = (\sqrt{3} – 1)\text{BD}\) \(\text{BD} = \dfrac{20}{\sqrt{3} – 1} \times \color{blue} \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}\) \(\text{BD} = \dfrac{20(\sqrt{3} + 1)}{3 – 1}\) \(\text{BD} = \dfrac{20\sqrt{3} + 20)}{2}\) \(\text{BD} = 10\sqrt{3} + 10 \text{ m}\) \(\text{AB} = \sqrt{3}\cdot\text{BD}\) \(\text{AB} = \sqrt{3}(10\sqrt{3} + 10)\) \(\text{AB} = 30 + 10\sqrt{3}\text{ m}\) Jadi tinggi tower tersebut adalah \(30 + 10\sqrt{3}\text{ m}\)