仅能使用初等几何原理得出角X的值。并写出详细步骤。
这道题是上一题的变形，但是要解出来也绝非易事。
基础几何原理
关于解题你所能用的原理。
相交线和交角：两条相交线，对角相等，两个邻角相加为180度。两条平行线与第三条线相交，相对的角相等。
三角形：三角形的内角总和是180度。等腰三角形有两条边相等，且有两个相等角。等边三角形三条边相等，三个角相等。直角三角形有一个九十度的角。两个三角形的形状相似，则它们的对应角相等。
∠AEB=180-60-50-30=40° ∠BDE=∠AED+10° ∠BDE+∠AED=180°-10°-40°-30° ∠AED=50° |
∠AEB=180-60-50-30=40° ∠BDE=∠AED+10° ∠BDE+∠AED=180°-10°-40°-30° ∠AED=50° |
第一步 ∠AEB=180-60-50-30=40° 好理解 第二步 ∠BDE=∠AED+10° 为什么？能否给出解释？ 第三步 ∠BDE+∠AED=180°-10°-40°-30° 好像不对吧，应该是 ∠BDE+∠AED=180°- 40°-30°= 110°
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ADB = 180 – 130 = 50 AEB = 180 – 140 = 40 AED+40+30 = BDE BDE = 180 – 20 – 50 – AED AED = 20 |
AED+40+30 = BDE 能否解释 |
ADB = 180 – 130 = 50 AEB = 180 – 140 = 40 AED+40+30 + BDE = 180 BDE = 180 – 20 – 50 – AED = 110 - AED BED = 180 – 30 - BDE BDE = 150 – BED = 150 – 110 + AED = 40 + AED 110 – AED = 40 + AED AED = 35 |
BDE = 110 - AED in line 6 you have 150 – BED = 150 – 110 + AED which implies BED=110-AED From the picture, you know that BDE and BED are two different angles. They may not be equal. You need to prove in line 6 150 – BED = 150 – 110 + AED |
let angle DEC = w, EDB = y and EDC = z then we have x+y = 110 w+z = 160 y+z = 130 x+w = 140 sovle this sytem of equations, the answers are w = 95 x = 45 y = 65 z = 65 |
This system has infinitely many solutions. You need to have more constrains. |
滴滴答答 |
实在不想再花2分钟写过程。 |
观察图片，会发现当所有已知角的度数确定以后，x的大小一定是确定的。所以给出无数的解或者列出一个无数解的方程式但只求出一解的都是不对的。一定有暗藏的条件我们还没有用到。不要忘了，这是世界上第二难的几何题 |
I don't know how to draw up here. step 1. draw a line from E to the side AE intersecting at point F, such that Step 2. triangle CFE and triangle ADE are congruent by ASA, so CF=AD Step 3. AD=AB, because step 4.draw a line from A to the side CB intersecting at point G, such that and Step 5. It can becocluded that www.ddhw.com |
Use the following formula:[edit] Law of cosinesThe law of cosines (also known as the cosine formula) is an extension of the Pythagorean theorem: also known as: In this formula the angle at C is opposite to the side c. This theorem can be proven by dividing the triangle into two right ones and using the Pythagorean theorem. The law of cosines is mostly used to determine a side of a triangle if two sides and an angle are known, although in some cases there can be two positive solutions as in the SSA ambiguous case. And can also be used to find the cosine of an angle (and consequently the angle itself) if all the sides are known. [edit] Other useful properties |
Use the following formular ..... then you can find the result Law of cosines The law of cosines (also known as the cosine formula) is an extension of the Pythagorean theorem: also known as: |
you are right. i mean to line AC . wish i can post the drawing |
Suppose the length of AB = 1 unit Because Angle ADB = Angle ABD = 50, so AD= 1 unit Angle AEB =40 According to Sine law: AE/sin80 = AB/sin40= 1 unit /sin(40), so AE=1unit * sin80/sin40 We now know the length of AD and AE, and Angle DAE=20, according to cosine law: DE^2=AD^2+AE^2-2*AD*AE*cos(20) After we get DE, we can compute x based on sine law AD/sin (x) = DE/sin(20), so sin(x) = sin(20)*AD/DE, then x can be computed. |
A typo: 'sine law' should be 'sine rule' and 'cosine law' should be 'cosine rule' |
if I am not wrong. too many solusions. e.g if α=-10，then β=170，then γ=-40，then x=150。this is impossible. |
could you give a simple solusion? |
Hint: Try to prove line BD = BE first. You may need some assistant lines drawn to provide more trangles. My answer is 25. |
Sorry, I meant 35. |
哈哈 终于找到和我的解法和答案一样的了 |
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