Prove the Extended Law of Sines

  • Thread starter Thread starter lema21
  • Start date Start date
  • Tags Tags
    Law Law of sines
lema21
Messages
18
Reaction score
9
Homework Statement
Prove the extended Law of Sines.
Hint: Let gamma be the circumscribed circle of triangle ABC and let D be the point on gamma such that DB is a diameter of gamma. Prove that Angle BAC is congruent to Angle BDC. Use that result to prove that sin(angle BAC) = BC/2R, where R is the circumradius.
Relevant Equations
Extended Law of sines: [(BC)/sin(angle BAC)] = [AC/sin(angle ABC)] = [AB/sin(angle ACB)] = 2r
I just want to know if this proof is okay, and I would like advise on how to improve it.
Proof:
i. Proving that ∠BAC ⩭∠BDC:
Let gamma be the circumscribed circle of ABC.
Let D be the point on gamma such that DB is a diameter of gamma.
The sum of the angles within a triangle equal to 180°.
Adding the angles of BAC together, b+a+c =180° where b=∠B, a=∠A, and c=∠C.
Adding the angles of BDC together, b+d+c = 180°, where b=∠B, d=∠D, and c=∠C.
Since the sum of both BAC and BDC equals 180°, b+a+c=b+d+c → a=d.
Hence, ∠BAC ⩭∠BDC.
ii. Proving that sin(∠BAC)=BC/2R:
sin(∠BAC) = P/H
= BC/BD
= BC/sin(∠BDC)
= BD
= 2R
sin(∠ADB) = P/H
= AB/BD
= AB/sin(∠ADB)
= BD
= 2R
AC/sin(∠ABC)= 2R
(BC/sin(∠BAC)) = (AC/sin(∠ABC) = (AB/sin(∠ACB)) = 2R
 
Physics news on Phys.org
\angle BOC=2\angle BAC = 2\angle BDC
may be helpful.
 
  • Like
Likes Charles Link
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top