Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
Trong to\u00e1n h\u00ecnh h\u1ecdc, t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn l\u00e0 m\u1ed9t trong nh\u1eefng ki\u1ebfn th\u1ee9c c\u1ea7n \u0111\u1eb7c bi\u1ec7t n\u1eafm ch\u1eafc. \u0110i\u1ec1u n\u00e0y l\u00e0 t\u01b0\u01a1ng \u0111\u1ed1i d\u1ec5 hi\u1ec3u khi n\u00f3 g\u00f3p m\u1eb7t \u1edf r\u1ea5t nhi\u1ec1u c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp kh\u00e1c nhau. V\u1eady c\u1ee5 th\u1ec3, t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn l\u00e0 g\u00ec? C\u00f3 nh\u1eefng ki\u1ebfn th\u1ee9c n\u00e0o c\u1ea7n nh\u1edb v\u1ec1 t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn?<\/p>\n\n\n\n
Gi\u1ea3i \u0111\u00e1p t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn<\/p>\n\n\n\n
T\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn l\u00e0 g\u00ec?<\/p>\n\n\n\n
T\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn c\u1ee5 th\u1ec3 l\u00e0 c\u00e1c t\u1ef7 s\u1ed1 v\u1ec1 c\u1ea1nh c\u1ee7a g\u00f3c nh\u1ecdn xu\u1ea5t hi\u1ec7n trong c\u00e1c tam gi\u00e1c vu\u00f4ng. Cho tam gi\u00e1c vu\u00f4ng ABC vu\u00f4ng t\u1ea1i A, g\u00f3c C l\u00e0 g\u00f3c nh\u1ecdn \u0111\u01b0\u1ee3c k\u00ed hi\u1ec7u l\u00e0 \u03b1.<\/p>\n\n\n\n
Ta c\u00f3: sin \u03b1= c\u1ea1nh \u0111\u1ed1i\/ c\u1ea1nh huy\u1ec1n = AB\/BC<\/p>\n\n\n\n
cos \u03b1= c\u1ea1nh k\u1ec1\/c\u1ea1nh huy\u1ec1n = AC\/BC<\/p>\n\n\n\n
tan \u03b1= c\u1ea1nh \u0111\u1ed1i\/c\u1ea1nh k\u1ec1 = AB\/AC<\/p>\n\n\n\n
cot \u03b1= c\u1ea1nh k\u1ec1\/c\u1ea1nh \u0111\u1ed1i = AC\/AB<\/p>\n\n\n\n
M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa t\u1ef7 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a 2 g\u00f3c nh\u1ecdn trong tam gi\u00e1c vu\u00f4ng<\/p>\n\n\n\n
Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. G\u00f3c B v\u00e0 g\u00f3c C ch\u00ednh l\u00e0 2 g\u00f3c nh\u1ecdn c\u1ee7a tam gi\u00e1c vu\u00f4ng v\u00e0 \u0111\u1ed3ng th\u1eddi ph\u1ee5 nhau. Ta c\u00f3 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a 2 g\u00f3c n\u00e0y nh\u01b0 sau:<\/p>\n\n\n\n
cosB = sin C; sinB = cosC; tanB = cotC; cotB = tanC.<\/p>\n\n\n\n
M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa c\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c<\/p>\n\n\n\n
Nh\u1eefng t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c \u0111\u1eb7c bi\u1ec7t trong tam gi\u00e1c vu\u00f4ng<\/p>\n\n\n\n
Trong tam gi\u00e1c vu\u00f4ng c\u00f3 m\u1ed9t s\u1ed1 nh\u1eefng g\u00f3c \u0111\u1eb7c bi\u1ec7t. \u0110i k\u00e8m v\u1edbi \u0111\u00f3 c\u0169ng l\u00e0 nh\u1eefng t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c \u0111\u1eb7c bi\u1ec7t. B\u1ea1n \u0111\u1ecdc c\u00f3 th\u1ec3 tham kh\u1ea3o \u1edf b\u1ea3ng t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c \u0111\u1eb7c bi\u1ec7t d\u01b0\u1edbi \u0111\u00e2y:<\/p>\n\n\n\n
0\u2070 30\u2070 45\u2070 60\u2070 90\u2070<\/p>\n\n\n\n
Sin 0 1\/2 \u221a2\/2 \u221a3\/2 1<\/p>\n\n\n\n
Cos 1 \u221a3\/2 \u221a2\/2 1\/2 0<\/p>\n\n\n\n
Tan 0 \u221a3\/3 1 \u221a3 Kh\u00f4ng x\u00e1c \u0111\u1ecbnh<\/p>\n\n\n\n
Cot Kh\u00f4ng x\u00e1c \u0111\u1ecbnh \u221a3 1 1\/\u221a3 0<\/p>\n\n\n\n
C\u00e1c d\u1ea1ng to\u00e1n v\u1ec1 t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c l\u1edbp 9 th\u01b0\u1eddng g\u1eb7p<\/p>\n\n\n\n
C\u00f3 3 d\u1ea1ng to\u00e1n v\u1ec1 c\u00f4ng th\u1ee9c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c l\u1edbp 9 th\u01b0\u1eddng g\u1eb7p nh\u1ea5t bao g\u1ed3m:<\/p>\n\n\n\n
T\u00ednh to\u00e1n t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c, c\u00e1c c\u1ea1nh, c\u00e1c g\u00f3c: S\u1eed d\u1ee5ng c\u00e1c c\u00f4ng th\u1ee9c v\u1ec1 sin, cos, tan v\u00e0 cot \u0111\u1ec3 l\u00e0m b\u00e0i. Ghi nh\u1edb c\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c g\u00f3c \u0111\u1eb7c bi\u1ec7t ph\u1ee5c v\u1ee5 t\u00ednh to\u00e1n.<\/p>\n\n\n\n
So s\u00e1nh c\u00e1c g\u00f3c d\u1ef1a v\u00e0o t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c: B\u00ean c\u1ea1nh c\u00e1ch t\u00ednh to\u00e1n b\u1eb1ng s\u1ed1 th\u00ec ng\u01b0\u1eddi l\u00e0m c\u00f3 th\u1ec3 \u0111\u01b0a c\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c v\u1ec1 c\u00f9ng lo\u1ea1i \u0111\u1ec3 so s\u00e1nh<\/p>\n\n\n\n
R\u00fat g\u1ecdn c\u00e1c bi\u1ec3u th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c: D\u1ef1a v\u00e0o c\u00e1c c\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c \u0111\u1eb7c bi\u1ec7t \u0111\u1ec3 r\u00fat g\u1ecdn c\u00e1c bi\u1ec3u th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c.<\/p>\n\n\n\n
C\u00e1c d\u1ea1ng to\u00e1n v\u1ec1 t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c th\u01b0\u1eddng g\u1eb7p<\/p>\n\n\n\n
B\u1ea3ng l\u01b0\u1ee3ng gi\u00e1c<\/p><\/blockquote>\n\n\n\n
B\u00e0i t\u1eadp t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c trong g\u00f3c nh\u1ecdn tr\u1eafc nghi\u1ec7m<\/p>\n\n\n\n
Ngay sau \u0111\u00e2y, h\u00e3y c\u00f9ng Vietlearn \u0111i gi\u1ea3i m\u00e3i m\u1ed9t v\u00e0i nh\u1eefng b\u00e0i to\u00e1n 9 t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn.<\/p>\n\n\n\n
C\u00e2u 1: C\u00f3 m\u1ea5y t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn<\/p>\n\n\n\n
4<\/p>\n\n\n\n
3<\/p>\n\n\n\n
5<\/p>\n\n\n\n
6<\/p>\n\n\n\n
C\u00f3 t\u1ed5ng c\u1ed9ng 4 t\u1ef7 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c trong g\u00f3c nh\u1ecdn c\u1ee7a tam gi\u00e1c vu\u00f4ng bao g\u1ed3m: sin, cos, tan v\u00e0 cot.<\/p>\n\n\n\n
Ch\u1ecdn A.<\/p>\n\n\n\n
C\u00e2u 2: Theo l\u00fd thuy\u1ebft t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn, sin 60\u2070 b\u1eb1ng bao nhi\u00eau?<\/p>\n\n\n\n
\u221a3\/2<\/p>\n\n\n\n
1\/2<\/p>\n\n\n\n
\u221a3\/3<\/p>\n\n\n\n
\u221a2\/2<\/p>\n\n\n\n
Ch\u1ecdn A.<\/p>\n\n\n\n
C\u00e2u 3: Tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. C\u00f3 g\u00f3c B b\u1eb1ng 45\u2070. Sin C b\u1eb1ng bao nhi\u00eau?<\/p>\n\n\n\n
\u221a3\/2<\/p>\n\n\n\n
\u221a2\/2<\/p>\n\n\n\n
1\/2<\/p>\n\n\n\n
1\/3<\/p>\n\n\n\n
G\u00f3c B b\u1eb1ng 45\u2070 suy ra g\u00f3c C b\u1eb1ng 45\u2070. Sin 45\u2070 = \u221a2\/2.<\/p>\n\n\n\n
Ch\u1ecdn B.<\/p>\n\n\n\n
C\u00e2u 4: sin2 \u03b1 + cos2 \u03b1 b\u1eb1ng bao nhi\u00eau?<\/p>\n\n\n\n
1<\/p>\n\n\n\n
2<\/p>\n\n\n\n
5<\/p>\n\n\n\n
3<\/p>\n\n\n\n
Ch\u1ecdn A<\/p>\n\n\n\n
C\u00e2u 5: Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang?<\/p>\n\n\n\n
sin B = cos C<\/p>\n\n\n\n
tan B = cos C<\/p>\n\n\n\n
tan B = sin C<\/p>\n\n\n\n
cos B = cot C<\/p>\n\n\n\n
Trong tam gi\u00e1c vu\u00f4ng ABC c\u00f3: sin B = cos C; cos B = sin C; tan B = cot C, cot B = tan C.<\/p>\n\n\n\n
Ch\u1ecdn A.<\/p>\n\n\n\n
C\u00e2u 6: Cho tam gi\u00e1c vu\u00f4ng ABC c\u00f3 AB = 3, AC = 4 v\u00e0 BC = 5. T\u00ednh cos C v\u00e0 tan B.<\/p>\n\n\n\n
cos C = 1\/2 ; tan B = \u221a3\/2<\/p>\n\n\n\n
cos C = 4\/5 ; tan B = 4\/3<\/p>\n\n\n\n
cos C = 3\/5 ; tan B = 4\/3<\/p>\n\n\n\n
cos C = 4\/5 ; tan B = 3\/4<\/p>\n\n\n\n
Ta c\u00f3: cos C = AC\/BC = 4\/5 ; tan B = AC\/AB = 4\/3.<\/p>\n\n\n\n
Ch\u1ecdn B.<\/p>\n\n\n\n
C\u00e2u 7: Nh\u1eadn \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y ch\u00ednh x\u00e1c.<\/p>\n\n\n\n
sin2x + cos2x = 2<\/p>\n\n\n\n
cos2x = 1+ tan2x<\/p>\n\n\n\n
sin2x = 1 + cot2x<\/p>\n\n\n\n
tanx.cotx = 1<\/p>\n\n\n\n
Tanx.cotx = (sinx\/cosx). (cosx\/sinx) = 1.<\/p>\n\n\n\n
Ch\u1ecdn D.<\/p>\n\n\n\n
C\u00e2u 8: Cho tan A = 5\/12. H\u1ecfi cos A b\u1eb1ng bao nhi\u00eau?<\/p>\n\n\n\n
12\/15<\/p>\n\n\n\n
11\/13<\/p>\n\n\n\n
14\/13<\/p>\n\n\n\n
12\/13<\/p>\n\n\n\n
Ta c\u00f3: 1\/(cos2A) = 1 + tan2 A \u2b04 1\/(cos2A) = 1+(5\/12)2 \u2b04 cos2A = 144\/169 \u2b04<\/p>\n\n\n\n
Cos A = 12\/13.<\/p>\n\n\n\n
Ch\u1ecdn D.<\/p>\n\n\n\n
C\u00e2u 9: Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. K\u1ebb \u0111\u01b0\u1eddng trung tuy\u1ebfn AM. Bi\u1ebft AB = 4, cosB = 1\/2. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n AM.<\/p>\n\n\n\n
5<\/p>\n\n\n\n
6<\/p>\n\n\n\n
4<\/p>\n\n\n\n
3<\/p>\n\n\n\n
Ta c\u00f3 tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. cosB = AB\/BC \u2b04 1\/2 = 4\/BC \u2b04 BC = 8. L\u1ea1i c\u00f3 AM l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n n\u00ean AM b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n. T\u1ee9c l\u00e0 AM = 1\/2 BC = 4.<\/p>\n\n\n\n
Ch\u1ecdn C.<\/p>\n\n\n\n
C\u00e2u 10: Cho tam gi\u00e1c MNP vu\u00f4ng t\u1ea1i M, \u0111\u01b0\u1eddng cao MI. G\u00f3c P b\u1eb1ng 60 \u0111\u1ed9, \u0111\u1ed9 d\u00e0i c\u1ea1nh MP b\u1eb1ng 4cm. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng cao MI.<\/p>\n\n\n\n
2\u221a5 cm<\/p>\n\n\n\n
2\u221a3 cm<\/p>\n\n\n\n
3\u221a3 cm<\/p>\n\n\n\n
1\/2 cm<\/p>\n\n\n\n
MI l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean MI vu\u00f4ng g\u00f3c v\u1edbi NP. Tam gi\u00e1c MIP vu\u00f4ng t\u1ea1i I. Ta c\u00f3 g\u00f3c P b\u1eb1ng 60 \u0111\u1ed9, \u0111\u1ed9 d\u00e0i c\u1ea1nh MP b\u1eb1ng 4cm. SinP = MI\/MP \u2b04 sin 60\u2070 = MI\/4<\/p>\n\n\n\n
=> MI = 2\u221a3<\/p>\n\n\n\n
Ch\u1ecdn B.<\/p>\n\n\n\n
Tr\u00ean \u0111\u00e2y l\u00e0 10 b\u00e0i to\u00e1n v\u1ec1 t\u1ef7 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn c\u01a1 b\u1ea3n \u0111\u1ec3 b\u1ea1n \u0111\u1ecdc c\u00f3 th\u1ec3 tr\u1ea3i nghi\u1ec7m. Hy v\u1ecdng th\u00f4ng qua b\u00e0i vi\u1ebft, c\u00e1c b\u1ea1n s\u1ebd c\u00f3 c\u00e1i nh\u00ecn t\u1ed5ng qu\u00e1t h\u01a1n v\u1ec1 ch\u1ee7 \u0111\u1ec1 t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn l\u1edbp 9. \u0110\u1ec3 h\u1ecdc t\u1ed1t to\u00e1n h\u00ecnh h\u1ecdc c\u00f2n ch\u1ea7n ch\u1edd g\u00ec m\u00e0 kh\u00f4ng theo d\u00f5i Vietlearn.<\/p>\n\n\n\n
T\u00ecm hi\u1ec3u th\u00eam:<\/p>\n\n\n\n
C\u0103n b\u1eadc ba<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":"
Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn Trong to\u00e1n h\u00ecnh h\u1ecdc, t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn l\u00e0 m\u1ed9t trong nh\u1eefng ki\u1ebfn th\u1ee9c c\u1ea7n \u0111\u1eb7c bi\u1ec7t n\u1eafm ch\u1eafc. \u0110i\u1ec1u n\u00e0y l\u00e0 t\u01b0\u01a1ng \u0111\u1ed1i d\u1ec5 hi\u1ec3u khi n\u00f3 g\u00f3p m\u1eb7t \u1edf r\u1ea5t nhi\u1ec1u c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp kh\u00e1c nhau. V\u1eady c\u1ee5 th\u1ec3, t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_disable_autopaging":false},"categories":[9],"tags":[],"class_list":["post-1085","post","type-post","status-publish","format-standard","hentry","category-toan-hoc"],"_links":{"self":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/1085","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/comments?post=1085"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/1085\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=1085"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=1085"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=1085"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}