{"id":620707,"date":"2026-11-09T09:35:37","date_gmt":"2026-11-09T02:35:37","guid":{"rendered":"https:\/\/hoanghamobile.com\/tin-tuc\/?p=620707"},"modified":"2026-11-09T09:35:37","modified_gmt":"2026-11-09T02:35:37","slug":"tam-giac-nhon","status":"publish","type":"post","link":"https:\/\/hoanghamobile.com\/tin-tuc\/tam-giac-nhon\/","title":{"rendered":"Tam gi\u00e1c nh\u1ecdn l\u00e0 g\u00ec? C\u00e1ch nh\u1eadn bi\u1ebft, t\u00ednh ch\u1ea5t v\u00e0 b\u00e0i t\u1eadp"},"content":{"rendered":"

Tam gi\u00e1c nh\u1ecdn l\u00e0 tam gi\u00e1c c\u00f3 c\u1ea3 ba g\u00f3c \u0111\u1ec1u l\u00e0 g\u00f3c nh\u1ecdn, t\u1ee9c m\u1ed7i g\u00f3c nh\u1ecf h\u01a1n 90\u00b0. Khi bi\u1ebft \u0111\u1ed9 d\u00e0i ba c\u1ea1nh a, b, c (c l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t), tam gi\u00e1c l\u00e0 tam gi\u00e1c nh\u1ecdn n\u1ebfu c\u00b2 < a\u00b2 + b\u00b2. Tam gi\u00e1c \u0111\u1ec1u lu\u00f4n l\u00e0 tam gi\u00e1c nh\u1ecdn. B\u00e0i vi\u1ebft gi\u1ea3i th\u00edch \u0111\u1ecbnh ngh\u0129a, c\u00e1ch nh\u1eadn bi\u1ebft, t\u00ednh ch\u1ea5t c\u00e1c \u0111\u01b0\u1eddng \u0111\u1eb7c bi\u1ec7t, c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch v\u00e0 b\u00e0i t\u1eadp c\u00f3 l\u1eddi gi\u1ea3i v\u1ec1 tam gi\u00e1c nh\u1ecdn.<\/p>\n

Tam gi\u00e1c nh\u1ecdn l\u00e0 g\u00ec?<\/h2>\n

Tam gi\u00e1c nh\u1ecdn l\u00e0 tam gi\u00e1c c\u00f3 ba g\u00f3c trong \u0111\u1ec1u l\u00e0 g\u00f3c nh\u1ecdn (l\u1edbn h\u01a1n 0\u00b0 v\u00e0 nh\u1ecf h\u01a1n 90\u00b0). V\u00ec t\u1ed5ng ba g\u00f3c c\u1ee7a m\u1ed9t tam gi\u00e1c lu\u00f4n b\u1eb1ng 180\u00b0, ch\u1ec9 c\u1ea7n g\u00f3c l\u1edbn nh\u1ea5t nh\u1ecf h\u01a1n 90\u00b0 l\u00e0 c\u1ea3 ba g\u00f3c \u0111\u1ec1u nh\u1ecdn.<\/p>\n

V\u00ed d\u1ee5: tam gi\u00e1c c\u00f3 ba g\u00f3c 50\u00b0, 60\u00b0, 70\u00b0 l\u00e0 tam gi\u00e1c nh\u1ecdn; tam gi\u00e1c c\u00f3 c\u00e1c g\u00f3c 30\u00b0, 60\u00b0, 90\u00b0 l\u00e0 tam gi\u00e1c vu\u00f4ng, kh\u00f4ng ph\u1ea3i tam gi\u00e1c nh\u1ecdn.\"tam-giac-nhon-1\"<\/p>\n

Ph\u00e2n lo\u1ea1i tam gi\u00e1c theo g\u00f3c<\/h2>\n\n\n\n\n\n\n\n
Lo\u1ea1i tam gi\u00e1c<\/th>\n\u0110\u1eb7c \u0111i\u1ec3m g\u00f3c<\/th>\nD\u1ea5u hi\u1ec7u theo c\u1ea1nh (c l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t)<\/th>\n<\/tr>\n<\/thead>\n
Tam gi\u00e1c nh\u1ecdn<\/td>\nC\u1ea3 3 g\u00f3c < 90\u00b0<\/td>\nc\u00b2 < a\u00b2 + b\u00b2<\/td>\n<\/tr>\n
Tam gi\u00e1c vu\u00f4ng<\/td>\nC\u00f3 1 g\u00f3c = 90\u00b0<\/td>\nc\u00b2 = a\u00b2 + b\u00b2<\/td>\n<\/tr>\n
Tam gi\u00e1c t\u00f9<\/td>\nC\u00f3 1 g\u00f3c > 90\u00b0<\/td>\nc\u00b2 > a\u00b2 + b\u00b2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

D\u1ea5u hi\u1ec7u theo c\u1ea1nh \u0111\u01b0\u1ee3c suy ra t\u1eeb \u0111\u1ecbnh l\u00fd cosin<\/a>: c\u00b2 = a\u00b2 + b\u00b2 \u2212 2ab\u00b7cosC, n\u00ean g\u00f3c C nh\u1ecdn khi v\u00e0 ch\u1ec9 khi cosC > 0, t\u1ee9c c\u00b2 < a\u00b2 + b\u00b2.<\/p>\n

Ngo\u00e0i ph\u00e2n lo\u1ea1i theo g\u00f3c, tam gi\u00e1c c\u00f2n \u0111\u01b0\u1ee3c ph\u00e2n lo\u1ea1i theo c\u1ea1nh (th\u01b0\u1eddng, c\u00e2n, \u0111\u1ec1u). M\u1ed9t tam gi\u00e1c c\u00f3 th\u1ec3 v\u1eeba nh\u1ecdn v\u1eeba c\u00e2n, v\u00ed d\u1ee5 tam gi\u00e1c c\u00e2n c\u00f3 g\u00f3c \u1edf \u0111\u1ec9nh b\u1eb1ng 40\u00b0.<\/p>\n

C\u00e1ch nh\u1eadn bi\u1ebft tam gi\u00e1c nh\u1ecdn<\/h2>\n

C\u00e1ch 1: D\u1ef1a v\u00e0o s\u1ed1 \u0111o g\u00f3c<\/h3>\n

T\u00ecm g\u00f3c l\u1edbn nh\u1ea5t c\u1ee7a tam gi\u00e1c. N\u1ebfu g\u00f3c l\u1edbn nh\u1ea5t nh\u1ecf h\u01a1n 90\u00b0 th\u00ec tam gi\u00e1c l\u00e0 tam gi\u00e1c nh\u1ecdn.<\/p>\n

\"tam-giac-nhon-2\"<\/p>\n

C\u00e1ch 2: D\u1ef1a v\u00e0o \u0111\u1ed9 d\u00e0i ba c\u1ea1nh<\/h3>\n
    \n
  1. X\u00e1c \u0111\u1ecbnh c\u1ea1nh l\u1edbn nh\u1ea5t, g\u1ecdi l\u00e0 c.<\/li>\n
  2. T\u00ednh c\u00b2 v\u00e0 a\u00b2 + b\u00b2.<\/li>\n
  3. So s\u00e1nh: n\u1ebfu c\u00b2 < a\u00b2 + b\u00b2 th\u00ec tam gi\u00e1c nh\u1ecdn.<\/li>\n<\/ol>\n

    V\u00ed d\u1ee5 1: Tam gi\u00e1c c\u00f3 ba c\u1ea1nh 5, 6, 7. C\u1ea1nh l\u1edbn nh\u1ea5t l\u00e0 7: 7\u00b2 = 49; 5\u00b2 + 6\u00b2 = 61. V\u00ec 49 < 61 n\u00ean \u0111\u00e2y l\u00e0 tam gi\u00e1c nh\u1ecdn (g\u00f3c l\u1edbn nh\u1ea5t x\u1ea5p x\u1ec9 78,5\u00b0).<\/p>\n

    V\u00ed d\u1ee5 2: Tam gi\u00e1c c\u00f3 ba c\u1ea1nh 4, 5, 7: 7\u00b2 = 49; 4\u00b2 + 5\u00b2 = 41. V\u00ec 49 > 41 n\u00ean \u0111\u00e2y l\u00e0 tam gi\u00e1c t\u00f9 (g\u00f3c l\u1edbn nh\u1ea5t x\u1ea5p x\u1ec9 101,5\u00b0).<\/p>\n

    V\u00ed d\u1ee5 3: Tam gi\u00e1c c\u00f3 ba c\u1ea1nh 6, 8, 10: 10\u00b2 = 100 = 6\u00b2 + 8\u00b2, n\u00ean \u0111\u00e2y l\u00e0 tam gi\u00e1c vu\u00f4ng.<\/p>\n

    T\u00ednh ch\u1ea5t c\u1ee7a tam gi\u00e1c nh\u1ecdn<\/h2>\n