বর্ধিত সাইন সূত্র

সাইন ল এর মাধ্যমে আমরা কোন ত্রিভুজের বাহু ও কোণের মধ্যে সম্পর্ক খুজে পাই।

সাইন ল বলে, কোন ত্রিভুজ $\triangle ABC$ তে,

$\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin\angle ABC}=\frac{AB}{\sin\angle ACB}$

এখন আমরা দেখব সাইন ল এর বর্ধিত অংশ।

$\frac{BC}{\sin\angle BAC}=\frac{AC}{\sin\angle ABC}=\frac{AB}{\sin\angle ACB}=2R$

এখানে $R$ হলো $\triangle ABC$ এর পরিবৃত্তের ব্যাসার্ধ। আচ্ছা, তুমি কী এটি প্রমাণ করতে পারবে? একটু চিন্তা করে দেখো সূত্রটি নিয়ে।

তাহলে এর প্রমাণ টা দেখি। $\triangle ABC$ এর পরিবৃত্তের ওপর এমন একটি বিন্দু $D$ নেই যেন $BD$ পরিবৃত্তের ব্যাস হয়।

এখন,

$\frac{BC}{\sin\angle BAC}=\frac{BC}{\sin\angle BDC}$

এবং, যেহেতু $\angle BCD$ সমকোণ, $\frac{BC}{\sin\angle BAC}=\frac{BC}{\sin\angle BDC}=\frac{BC}{BC/BD}=BD=2R$

খুবই সুন্দর একটি প্রমাণ।

চল এখন বর্ধিত সাইন ল ব্যবহার করে একটি নতুন আইডেন্টিটি প্রমাণ করি। কোনো $\triangle ABC$ এর ক্ষেত্রফল $=\frac{abc}{4R}$। এখানে $a=BC, b=AC, c= AB$ এবং $R=$ পরিবৃত্তের ব্যাসার্ধ।

আমরা জানি, কোনো $\triangle ABC$ এর ক্ষেত্রফল $=\frac{1}{2}ab\cdot\sin \angle ACB$।

আবার, $\sin \angle ACB=\frac{c}{2R}$। এ দুটি একসঙ্গে করে আমরা আমাদের অভেদের প্রমাণ পাই।

The Law of Sines is:

$A \sin a+B\sin b+C\sin c=2r$

The other Law of Sines is $\frac{A}{\sin a}=\frac{B}{b\sin b}$, but with the same relation with B,C and A,C.

So the law of sines states that, for a triangle $ABC$ where the lower-case letter stands for the length of the side opposite the vertex with the corresponding capital letter (e.g. $a=BC$), we have that $$\frac {\sin \angle A}{a} = \frac {\sin \angle B}{b} = \frac {\sin \angle C}{c}.$$So we know that these three ratios are equal in value. But what is this value? That's what the extended law of sines gives us an idea about. Click Continue to reveal the formula.

Extended Law of Sines. In a triangle $ABC$ with $a=BC$, $b=AC$, and $c=AB$, we have $$\frac {\sin \angle A}{a} = \frac {\sin \angle B}{b} = \frac {\sin \angle C}{c} = \frac {1}{2R},$$where $R$ is the circumradius of the triangle.

Proof (Taken from the AoPSWiki.) In the diagram below, point $O$ is the circumcenter of $\triangle ABC$. Point $D$ is on $BC$ such that $\overline{OD}\perp\overline{BC}$. Since $\triangle ODB \cong \triangle ODC$, we have $$BD = CD = \frac {a}{2}$$ and $\angle BOD = \angle COD$. Furthermore, we have $\angle BOC = \angle 2 \angle BAC$, so $\angle BOD = \angle COD = \theta$. We can use trigonometry on right $\triangle BOD$ to find that $$\sin \theta = \frac {a}{2R} \implies \frac {a}{\sin \theta} = 2R.$$This can easily be done for $b$ and $c$ as well, so we are done. $\blacksquare$

Accompanying diagram.

The Law of Sines states that:

$\frac{A}{\sin a} + \frac{B}{\sin b} + \frac{C}{\sin c} = 2R$

where $A,B,C$ are the sides of the triangle, $a,b,c$ are the corresponding opposite angles, and $R$ is the circumradius.

This is very useful, with varied applications in mathematics.

The law of sines in triangle ABC, with angles a, b, and c across from A, B and C respectively, is:

$\frac{{\sin A}}{a} = \frac{{\sin B}}{b} = \frac{{\sin C}}{c} = 2R$

Where R is the circumradius of the triange.

The Law of Sines, or LoS, states that in a triangle with sides $a$, $b$, and $c$, $\dfrac{a}{\sin a} = \dfrac{b}{\sin b} = \dfrac{c}{\sin c} = 2R$

where R is the circumradius of ABC, or the radius of the circle that passes through all 3 vertices of the triangle.

$\huge{Proof:}$

For starters, we will only prove the first part of the LoS, namely $\dfrac{a}{\sin a} = \dfrac{b}{\sin b} = \dfrac{c}{\sin c}.$

If we call the sides BC, AC, and AB in the diagram $a$, $b$, and $c$ respectively in standard fashion and call the altitude dropped from A to D $h$, then we get by the definition of sine $$\begin {align*} \sin B &= \dfrac{h}{c}\\ \sin C &= \dfrac{h}{b}. \end {align*}$$ STOP here and think: Can i get it on my own from here? What should the next step be? THEN move on. :)

This simplifies into the desired: $$\begin {align} c \sin B &= h = b \sin C;\\ \dfrac {b}{\sin B} &= \dfrac {c}{\sin C}. \end {align}$$

Now, we can apply this to another side of the triangle to obtain the result including $\dfrac{a}{\sin A}.$

The second step is to prove that these are all equal to $2$ times $R$. However, this is easily done by realizing a strong connection between sines and the circumradius in a triangle.

More specifically, the area $\large{A}$ can be expressed in terms of both: $$\dfrac{ab\sin C}{2} = \large{A} = \dfrac{abc}{4R}.$$ Looking at this, we notice that we can simplify it (by simple division and rearrangement) to $$\begin {align} \dfrac{\sin C}{2} &= \dfrac{c}{4R}\\ \dfrac{\sin C}{c} &= \dfrac{2}{4R} = \dfrac{1}{2R}\\ \end {align}$$

Now we can just take the reciprocal of both sides, to receive $$\dfrac{c}{\sin C} = \dfrac{2R}{1} = 2R,$$

which is most definitely the result we wanted to see, and the complete proof of the Law of Sines $\dfrac{a}{\sin a} = \dfrac{b}{\sin b} = \dfrac{c}{\sin c} = 2R$ concludes. :D