tangents Law 如何求角?

p445hkk20001

(a + b) / (a - b) = tan((α+β)/2) / tan((α-β)/2)
when I know a,b,α   how to know β

-終場ソ使者-

Using compund angle formula
i.e. (a+b)/(a-b)={[tan(α/2)+tan(β/2)][1+tan(α/2)tan(β/2)]}/{[tan(α/2)-tan(β/2)][1-tan(α/2)tan(β/2)]}
β can be found easily

p445hkk20001

回復 2# -終場ソ使者-

Using compund angle formula
i.e. (a+b)/(a-b)={[tan(α/2)+tan(β/2)][1+tan(α/2)tan(β/2)]}/{[tan(α/ ...
-終場ソ使者- 發表於 1-4-2010 03:27 AM

but how can 主項轉換?
if a=8 b=12 α=36.33605751 how to know β?   β=117.2796127

-終場ソ使者-

(a+b)/(a-b)={[tan(α/2)+tan(β/2)][1+tan(α/2)tan(β/2)]}/{[tan(α/2)-tan(β/2)][1-tan(α/2)tan(β/2)]}
-5[tan(α/2)-tan(β/2)][1-tan(α/2)tan(β/2)]=[tan(α/2)+tan(β/2)][1+tan(α/2)tan(β/2)]
-5[tan(α/2)-tan²(α/2)tan(β/2)-tan(β/2)+tan²(β/2)tan(α/2)]=tan(α/2)+tan²(α/2)tan(β/2)+tan(β/2)+tan²(β/2)tan(α/2)
Let x,c be tan(β/2),tan(α/2) respectively,   (where c is a constant)
-5c+5c²x+5x-5cx²=c+c²x+x+cx²
6cx²-(4+4c²)x+6c=0
x={(4+4c²)±√[(4+4c²)²-4(6c)²]}/2
β=2tan^-1x

p445hkk20001

β=2tan-1x

Let x,c be tan(β/2),tan(α/2)

i don't understand .....
when you know β=2tan-1x 即 β=2tan-1[tan(α/2] 咁又點
so what?

講返我真正目的... 我係想用 知2角 1邊 黎知β
因為我想寫prog ,law of sine 係ASS 會有2個答案 so i really want to know laws of tangents can know the angle?

-終場ソ使者-

Law of tangents is an extra formula for calculating angles or edges,is useless,meaningless.
In fact,law of sines can have more than one anwers because of SINE.
For example
sin60°=sin120°
can we say 60°=sin^-1sin120° ?
The answer is,yes!
Since sinθ=sin(180°-θ) or sinθ=sin(π-θ)

p445hkk20001

做到呢個又點can we say 60°=sin-1sin120° ?
可以代表可以用law tangents 黎求角?

我都知sine law有2個答案

所以睇下tan law 可唔可以求角

-終場ソ使者-

算啦,law of tangents係廢架
你想搵兩個solutions用law of sines就夠啦
law of tangents真係好煩,
如果我冇計錯,β應該=2tan^-1{{(4+4c²)±√[(4+4c²)²-4(6c)²]}/2}

做到呢個又點can we say 60°=sin-1sin120° ?
可以代表可以用law tangents 黎求角?
呢兩句冇邏輯關係,我只係講law of sines可以有兩個solutions
況且中學既syllabus都冇law of tangents

p445hkk20001

算啦,law of tangents係廢架
你想搵兩個solutions用law of sines就夠啦
law of tangents真係好煩,
如果我冇 ...
-終場ソ使者- 發表於 2-4-2010 11:24 PM

我只係在想 方法 解決sine law 的2角答案

β應該=2tan-1{{(4+4c²)±√[(4+4c²)²-4(6c)²]}/2} 唔洗用a,b既資料?

[S]【YU】

真係辛苦晒 -終場ソ使者- 周旋咁耐,仲打埋個derivation俾佢.
sine law 都做倒既野,做咩要將簡單問題複雓化呢.
最重要係得出黎既general formula係一大抽野,有乜用.