Prove an eqaution on a triangle

[bilgunko]

If in a triangle ABC, angle A is 30°, angle B is 50° prove that c^2=b*(a+b).

Sides a, b, and c are opposite of angles A, B, and C respectively.

I think the problem should not be very hard as it was in the intermediate section of my math textbook. But I really can't see how to prove it.

I have been trying to solve this problem for a few hours. I've tried to draw altitudes from the angles and use the Pythagorian theorem on the right triangles that were formed. I've also tried drawing squares on the sides of the triangle and tried to equate the areas of the rectangles but I am not really sure whether it is right to do so.
I thought of finding identical triangles formed by the altitudes and prove some ratio but I don't know how to do it. I've actually asked my math teacher too but she couldn't prove the equation.

Here's the picture of the problem for convenience.

 

[bilgunko]

I just got this explanation from someone.

"Because
c^2 = a^2 + b^2 Pythagoras

= b(a^2/b + b) factor out b

This means that a^2/b has to be = a (from "to prove")

This would mean that a/b = 1. But this is only true in an isosceles
right triangle. So I think that what you're asked to prove is not true."

It seems to be wrong because when I substitute any number in one of the a,b or c sides the equation is proved.

 

[soroban]

Hello, bilgunko!

I found an approach, but it led me nowhere.
You probably tried it already, but if you haven't, maybe you can do something with it.

\(\displaystyle \text{If in }\Delta ABC,\:\angle A =30^o,\;\angle B = 50^o\text{, prove that: }\:c^2\:=\:b(a+b)\)

                              C 
                              *
                           *60|*
                        *     | *
                     *        |40*
              b   *           |   *  a
               *              |    *
            *                 |     *
         * 30                 |   50 *
    A * - - - - - - - - - - - * - - - * B
                              D

\(\displaystyle \text{We note that: }\,\Delta CDA\text{ is a 30-60 right triangle.}\)

. . \(\displaystyle \text{Hence: }\:CD = \tfrac{b}{2},\;\;AD = \tfrac{\sqrt{3}}{2}b \quad\Rightarrow\quad DB \:=\:c - \tfrac{\sqrt{3}}{2}b\)

\(\displaystyle \text{In right triangle }CDB\text{ we have: }\:\left(c - \tfrac{\sqrt{3}}{2}b\right)^2 + \left(\tfrac{b}{2}\right)^2 \:=\:a^2\)


But this gave me nothing like the desired result . . .

 

[galactus]

We have three angles. That is enough to solve for the sides.

Using the law of sines and the law of cosines and forming a system:

\(\displaystyle c^{2}+b^{2}-2abcos(100)\)

\(\displaystyle \frac{sin(30)}{a}=\frac{sin(100)}{c}\)

\(\displaystyle c=\frac{a\cdot sin(100)}{sin(30)}\)

\(\displaystyle \frac{sin(30)}{a}=\frac{sin(50)}{b}\Rightarrow b=\frac{a\cdot sin(50)}{sin(30)}\)

Sub into the law of cosines formula and we have it all in terms of a. Solving, gives us:

\(\displaystyle a=.5077133, \;\ b=.7778619343, \;\ c=1\)

\(\displaystyle c^{2}=b(a+b)=.7778619343(.5077133+.7778619343)=1^{2}=1\)

This method works OK.

 

[Denis]

If you let a=1 and use Law of Sines, then you need to prove:

[SIN(100) / SIN(30)]^2 = k(1 + k) where k = SIN(50) / SIN(30) ; remember that SIN(30) = 1/2

 

[bilgunko]

We have three angles. That is enough to solve for the sides.

Using the law of sines and the law of cosines and forming a system:

\(\displaystyle c^{2}+b^{2}-2abcos(100)\)

\(\displaystyle \frac{sin(30)}{a}=\frac{sin(100)}{c}\)

\(\displaystyle c=\frac{a\cdot sin(100)}{sin(30)}\)

\(\displaystyle \frac{sin(30)}{a}=\frac{sin(50)}{b}\Rightarrow b=\frac{a\cdot sin(50)}{sin(30)}\)

Sub into the law of cosines formula and we have it all in terms of a. Solving, gives us:

\(\displaystyle a=.5077133, \;\ b=.7778619343, \;\ c=1\)

\(\displaystyle c^{2}=b(a+b)=.7778619343(.5077133+.7778619343)=1^{2}=1\)

This method works OK.

Hi!
Thank you for all the help but I don't understand this sentence.
"Sub into the law of cosines formula and we have it all in terms of a. Solving, gives us:"

How are we getting the length of the sides in numbers when there were given only angles of the triangle?

 

[bilgunko]

Hi!
I just found the solution myself! Extending the line b by length of a and making an isosceles triangle we can see that the big triangle ABD (I forgot to denote the letter for point D but it's the only one left) is identical to ABC. From this we can write ratio.
(a+b)/c=c/b
thus
b*(a+b)=c^2

Thanks.