Forgot password
 Register account
View 1734|Reply 5

[几何] 三角形正弦余弦一题

[Copy link]

412

Threads

1432

Posts

3

Reputation

Show all posts

realnumber posted 2019-4-8 14:17 |Read mode
三角形ABC的三个角余弦值分别等于三角形DEF的三个角的正弦值,试判断它们的形状.
(选项:锐角三角形,直角三角形,钝角三角形)
这枚题目很有意思.

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2019-4-8 14:35
我记得很早以前就见过这题……

412

Threads

1432

Posts

3

Reputation

Show all posts

original poster realnumber posted 2019-4-8 14:43
应该很早就有的

7

Threads

578

Posts

9

Reputation

Show all posts

游客 posted 2019-4-8 21:17
三个余弦的显然是锐角三角形,那么另外一个必然是钝角三角形。

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2019-4-8 21:57
重新写一下,下午写得有点笨……
依题意有
\[
\cos A=\sin D=\cos(90\du-D)\riff A=\abs{90\du-D},
\]另外两者同理,故
\[
180\du=A+B+C=c_1(90\du-D)+c_2(90\du-E)+c_3(90\du-F),
\]其中 `c_i\in\{-1,1\}`,显然 `c_i` 不能全是 `1`,而取 `-1` 则意味着相应的角为钝角,但一个三角形内最多一个钝角,所以有且只有一个 `c_i` 取 `-1`,不妨设 `c_1=c_2=1`, `c_3=-1`,上式变成 $180\du=90\du-D-E+F=-90\du+2F$,解得 $F=135\du$。

综上得 `\triangle ABC` 有一角为 $45\du$,`\triangle DEF` 有一角为 $135\du$,其余两对角对应地互余。

412

Threads

1432

Posts

3

Reputation

Show all posts

original poster realnumber posted 2019-4-8 22:08
45,135度没想到,只是反证法证了下,
1>cosA=sinD>0,A为锐角,B,C一样
假设DEF都为锐角,那么$D=90\du -A$,EF一样,这和$D+E+F=A+B+C=180\du$
矛盾.此题外表好象难?唬住了一大批学生.

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-7-15 15:09 GMT+8

Powered by Discuz!

Processed in 0.012639 seconds, 22 queries