An riaghailt cosine
A' lorg taobh
'S e an riaghailt cosine:
\({a^2} = {b^2} + {c^2} - 2bcCosA\)
Cleachd am foirmle seo nuair a tha fios agad air dà thaobh agus an ceàrn eatarra.
Eisimpleir
Lorg faid BC.
Freagairt
Tha dà thaobh againn agus an ceàrn eatarra.
\({a^2} = {b^2} + {c^2} - 2bcCosA\)
\({a^2} = {7^2} + {3^2} - (2 \times 7 \times 3 \times \cos (35^\circ ))\)
\(a^{2}=49+9-34.40\)
\(a^{2}=23.60\)
\(a=\sqrt{23.60}\)
\(a = 4.9\,cm\,(gu\,1\,id)\)
A-nis feuch a' cheist gu h-ìosal.
Question
Lorg meud ceàrn AB.
Tha an dà thaobh againn agus an ceàrn eatarra.
\({c^2} = {a^2} + {b^2} - 2ab CosC\)
\({c^2} = {4^2} + {9^2} - 2 \times 4 \times 9 \times Cos27^\circ\)
\(c^2 = 16 + 81 - 64.15\)
\(c^2 = 32.85\)
\(c= \sqrt{32.85}\)
\(c = 5.7\,cm\)
Question
Lorg meud ceàrn AB.
(Am faic thu pàtran ann an litrichean an fhoirmle? Atharraich iad sin gus freagairt air a' cheist.)
\(cosB = \frac{{{{a^2} + {c^2} - {b^2}}}}{2ac}\)
\(cosB = \frac{{{{4^2} + {5^2} - {7^2}}}}{2 \times 4 \times 5}\)
\(cosB = \frac{{{{16} + {25} - {49}}}}{40}\)
\(cosB = \frac{-8}{40}\)
\(cosB = -0.2\)
\(\text{CeàrnB}\ = cos{^-}{^1}(-0.2)\)
\(\text{CeàrnB}\ = 101.5^\circ\)
Question
Lorg meud ceàrn R.
Seach gu bheil sinn ag obrachadh a-mach meud ceàirn, bidh sinn a' cleachdadh an dara foirmle.
\(\cos R = \frac{{{p^2} + {q^2} - {r^2}}}{{2pq}}\)
\(\cos R = \frac{{{{4.2}^2} + {{6.9}^2} - {4^2}}}{{2 \times 4.2 \times 6.9}}\)
\(\cos R = \frac{{49.25}}{{57.96}}\)
\(CosR=0.850\)
\(R=cos^{-1}(0.850)\)
\(R = 31.8^\circ (gu\,1\,id)\)