Take triangle ABC. Drop a perpendicular of height h from vertex C onto side AB.
Why this step? An altitude splits the triangle into two right triangles, and in a right triangle sin(angle)=hypotenuseopposite — that's the only trig fact we need.
In the right triangle on the left, angle A has opposite side h and hypotenuse b:
sinA=bh⟹h=bsinA
In the right triangle on the right, angle B has opposite side h and hypotenuse a:
sinB=ah⟹h=asinB
Why this step? Both expressions equal the same height h, so we can equate them.
bsinA=asinB⟹sinAa=sinBb
Repeat with an altitude from A onto BC to get sinBb=sinCc. Done for the ratio equality.
Compute sinB=absinA. The critical length is the altitude h=bsinA.
Condition
Number of triangles
a<h=bsinA
0 (side too short to reach base)
a=h
1 (right triangle, just touches)
h<a<b
2 (arc cuts base twice: B and 180°−B)
a≥b
1 (long side, only one valid B)
Why the "180°−B" second solution? Because sinB=sin(180°−B), the equation for B can have two answers in (0°,180°). Keep the second one only ifA+(180°−B)<180°, i.e. the angles still fit in a triangle.
Imagine a triangle drawn perfectly inside a circle. The Law of Sines says: take any side, divide it by the "sine" of the corner facing it — you always get the same number, and that number is exactly how wide the whole circle is (its diameter).
The tricky "ambiguous" part: suppose I tell you one corner's angle and two stick lengths, but not exactly how they connect. It's like swinging a door hinge — sometimes the door tip hits the wall in two spots, sometimes in one, sometimes it can't reach at all. So there can be two different triangles, one, or none!
Dekho, Law of Sines ka core idea bahut simple hai: kisi bhi triangle mein, jo side jitni badi hoti hai uska saamne wala angle bhi utna hi bada hota hai. Formula bolta hai sinAa=sinBb=sinCc — yaani side ko uske opposite angle ke sine se divide karo, teeno ke liye same number aayega. Aur mazedaar baat: yeh number exactly 2R hai, jahan R triangle ke bahar wale circle (circumcircle) ka radius hai. Proof mein hum altitude h girate hain, do right triangles bante hain, aur dono se h nikaal ke equate kar dete hain — bas ho gaya.
Ab asli twist hai ambiguous case (SSA). Jab tumhe do sides aur ek angle diya ho jo un dono ke beech mein nahi hai, tab problem hoti hai. Socho ek side ko darwaze ki tarah hinge se ghuma rahe ho — kabhi woh base line ko do jagah touch karti hai (do triangle), kabhi ek jagah, kabhi pahunchti hi nahi (koi triangle nahi). Critical length hai h=bsinA. Agar a<h toh triangle banega hi nahi, agar h<a<b toh do triangle banenge.
Do triangle kyun? Kyunki sinB=sin(180°−B) — matlab B ke do possible values ho sakte hain. Isliye jab bhi SSA problem aaye, calculator se mila B lo, phir 180°−B bhi check karo, aur dekho ki angle sum 180° se kam rehta hai ya nahi. Yeh cheez exams mein bahut marks kha jaati hai, toh hamesha dono solutions verify karo. Agar sinB>1 aa jaye, matlab triangle possible hi nahi — seedha "no solution" likh do.
