Buktikan rumus tersebutC² = √A²+B² - 2AB . Cos ∅

Buktikan rumus tersebutC² = √A²+B² – 2AB . Cos ∅

Penjelasan dengan langkah-langkah:

perhatikan gambar!

Pada segitiga BCD, definisi cosinus:

$cos(c) = frac{cd}{a}$

$cd = a cdot cos(c)$

sehingga panjang DA:

DA = b – cos (C)

dan

$sin(c) = frac{bd}{a}$

$a cdot sin(c) = bd$

seperti yang kita ketahui:

pada segitiga ADB berlaku aturan phytagoras:

${c}^{2} = {bd}^{2} + {da}^{2}$

${c}^{2} = (a cdot sin(c) )^{2} + (b - a cdot cos(c) )^{2}$

$small{{c}^{2} = {a}^{2} sin^{2} (c) + {b}^{2} - 2ab cdot cos(c) + {a}^{2} cdot cos^{2} (c) }$

$small{{c}^{2} = {a}^{2} cdot sin^{2} (c) + {a}^{2} cdot cos^{2} (c) + {b}^{2} - 2ab cdot cos(c) }$

$small{{c}^{2} = {a}^{2} (sin^{2} (c) + cos^{2} (c) )+ {b}^{2} - 2ab cdot cos(c) }$

terapkan identitas

$sin^{2} ( theta) + cos^{2} ( theta) = 1$

sehingga

$small{{c}^{2} = {a}^{2} (1 )+ {b}^{2} - 2ab cdot cos(c) }$

$small{{c}^{2} = {a}^{2}+ {b}^{2} - 2ab cdot cos(c) }$

terbukti!

Gambar Jawaban