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tolong bantu jawab jangan ngasal​

Berikut ini adalah pertanyaan dari mridho409 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Tolong bantu jawab jangan ngasal​
tolong bantu jawab jangan ngasal​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\cos{A}=\frac{2}{3},\,\,\,\,\cos{B}=\frac{3}{5}

Mencari perbandingan trigonometri sisi segitiga siku-siku:
(Ada pada gambar)

Maka,

a)

\cos{A}\,\cos{B}-\sin{A}\,\sin{B}\\\\=\frac{2}{3}\times\frac{3}{5}-\frac{\sqrt5}{3}\times\frac{4}{5}\\\\=\frac{2}{5}-\frac{4\sqrt5}{15}\\\\=\frac{6}{15}-\frac{4\sqrt5}{15}\\\\=\frac{6-4\sqrt5}{15}

b)

\sin{A}\,\cos{B}+\sin{B}\,\cos{A}\\\\=\frac{\sqrt5}{3}\times\frac{3}{5}+\frac{4}{5}\times\frac{2}{3}\\\\=\frac{\sqrt5}{5}+\frac{8}{15}\\\\=\frac{3\sqrt5}{15}+\frac{8}{15}\\\\=\frac{3\sqrt5+8}{15}

c)

\tan^2{A}-\tan^2{B}\\\\={(\frac{\sqrt5}{2})}^2-{(\frac{4}{3})}^2\\\\=\frac{5}{4}-\frac{16}{9}\\\\=\frac{45}{36}-\frac{64}{36}\\\\=\frac{19}{36}

d)

\sin^2{A}+\cos^2{B}\\\\={(\frac{\sqrt5}{3})}^2+{(\frac{3}{5})}^2\\\\=\frac{5}{9}+\frac{9}{25}\\\\=\frac{125}{225}+\frac{81}{225}\\\\=\frac{206}{225}

[tex]\cos{A}=\frac{2}{3},\,\,\,\,\cos{B}=\frac{3}{5}[/tex]Mencari perbandingan trigonometri sisi segitiga siku-siku:(Ada pada gambar)Maka,a)[tex]\cos{A}\,\cos{B}-\sin{A}\,\sin{B}\\\\=\frac{2}{3}\times\frac{3}{5}-\frac{\sqrt5}{3}\times\frac{4}{5}\\\\=\frac{2}{5}-\frac{4\sqrt5}{15}\\\\=\frac{6}{15}-\frac{4\sqrt5}{15}\\\\=\frac{6-4\sqrt5}{15}[/tex]b)[tex]\sin{A}\,\cos{B}+\sin{B}\,\cos{A}\\\\=\frac{\sqrt5}{3}\times\frac{3}{5}+\frac{4}{5}\times\frac{2}{3}\\\\=\frac{\sqrt5}{5}+\frac{8}{15}\\\\=\frac{3\sqrt5}{15}+\frac{8}{15}\\\\=\frac{3\sqrt5+8}{15}[/tex]c)[tex]\tan^2{A}-\tan^2{B}\\\\={(\frac{\sqrt5}{2})}^2-{(\frac{4}{3})}^2\\\\=\frac{5}{4}-\frac{16}{9}\\\\=\frac{45}{36}-\frac{64}{36}\\\\=\frac{19}{36}[/tex]d)[tex]\sin^2{A}+\cos^2{B}\\\\={(\frac{\sqrt5}{3})}^2+{(\frac{3}{5})}^2\\\\=\frac{5}{9}+\frac{9}{25}\\\\=\frac{125}{225}+\frac{81}{225}\\\\=\frac{206}{225}[/tex][tex]\cos{A}=\frac{2}{3},\,\,\,\,\cos{B}=\frac{3}{5}[/tex]Mencari perbandingan trigonometri sisi segitiga siku-siku:(Ada pada gambar)Maka,a)[tex]\cos{A}\,\cos{B}-\sin{A}\,\sin{B}\\\\=\frac{2}{3}\times\frac{3}{5}-\frac{\sqrt5}{3}\times\frac{4}{5}\\\\=\frac{2}{5}-\frac{4\sqrt5}{15}\\\\=\frac{6}{15}-\frac{4\sqrt5}{15}\\\\=\frac{6-4\sqrt5}{15}[/tex]b)[tex]\sin{A}\,\cos{B}+\sin{B}\,\cos{A}\\\\=\frac{\sqrt5}{3}\times\frac{3}{5}+\frac{4}{5}\times\frac{2}{3}\\\\=\frac{\sqrt5}{5}+\frac{8}{15}\\\\=\frac{3\sqrt5}{15}+\frac{8}{15}\\\\=\frac{3\sqrt5+8}{15}[/tex]c)[tex]\tan^2{A}-\tan^2{B}\\\\={(\frac{\sqrt5}{2})}^2-{(\frac{4}{3})}^2\\\\=\frac{5}{4}-\frac{16}{9}\\\\=\frac{45}{36}-\frac{64}{36}\\\\=\frac{19}{36}[/tex]d)[tex]\sin^2{A}+\cos^2{B}\\\\={(\frac{\sqrt5}{3})}^2+{(\frac{3}{5})}^2\\\\=\frac{5}{9}+\frac{9}{25}\\\\=\frac{125}{225}+\frac{81}{225}\\\\=\frac{206}{225}[/tex][tex]\cos{A}=\frac{2}{3},\,\,\,\,\cos{B}=\frac{3}{5}[/tex]Mencari perbandingan trigonometri sisi segitiga siku-siku:(Ada pada gambar)Maka,a)[tex]\cos{A}\,\cos{B}-\sin{A}\,\sin{B}\\\\=\frac{2}{3}\times\frac{3}{5}-\frac{\sqrt5}{3}\times\frac{4}{5}\\\\=\frac{2}{5}-\frac{4\sqrt5}{15}\\\\=\frac{6}{15}-\frac{4\sqrt5}{15}\\\\=\frac{6-4\sqrt5}{15}[/tex]b)[tex]\sin{A}\,\cos{B}+\sin{B}\,\cos{A}\\\\=\frac{\sqrt5}{3}\times\frac{3}{5}+\frac{4}{5}\times\frac{2}{3}\\\\=\frac{\sqrt5}{5}+\frac{8}{15}\\\\=\frac{3\sqrt5}{15}+\frac{8}{15}\\\\=\frac{3\sqrt5+8}{15}[/tex]c)[tex]\tan^2{A}-\tan^2{B}\\\\={(\frac{\sqrt5}{2})}^2-{(\frac{4}{3})}^2\\\\=\frac{5}{4}-\frac{16}{9}\\\\=\frac{45}{36}-\frac{64}{36}\\\\=\frac{19}{36}[/tex]d)[tex]\sin^2{A}+\cos^2{B}\\\\={(\frac{\sqrt5}{3})}^2+{(\frac{3}{5})}^2\\\\=\frac{5}{9}+\frac{9}{25}\\\\=\frac{125}{225}+\frac{81}{225}\\\\=\frac{206}{225}[/tex]

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Last Update: Tue, 03 May 22
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