finding the cosine

homeschool girl

Junior Member
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Feb 6, 2020
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113
The problem:

In triangle \(\displaystyle ABC\), \(\displaystyle AB = AC,\) and angle \(\displaystyle A\) is equal to \(\displaystyle 36^\circ.\) Point \(\displaystyle D\) is on \(\displaystyle \overline{AC}\) so that \(\displaystyle \overline{BD}\) bisects \(\displaystyle \angle ABC.\)


(a) Prove that \(\displaystyle BC = BD = AD\).


(b) Let \(\displaystyle x = BC\) and let \(\displaystyle y = CD\). Using similar triangles \(\displaystyle BCD\) and \(\displaystyle ABC\), write an equation involving \(\displaystyle x\) and \(\displaystyle y\).


(c) Let \(\displaystyle r = \frac{y}{x}\). Write the equation from Part (b) in terms of \(\displaystyle r,\) and find \(\displaystyle r.\)


(d) Find \(\displaystyle \cos 36^\circ\) and \(\displaystyle \cos 72^\circ\) using Parts a to c. (Do not use your calculator!)


here's what I have so far:

Part A

\(\displaystyle \triangle ABC\)is icocilese, so \(\displaystyle \angle ABC = \angle ACB = \frac{180-36}{2}=72\).


adding the angle bisector \(\displaystyle BD\), \(\displaystyle \angle DBC = \frac{72}{2} = 36\).


So \(\displaystyle \triangle ABC \sim \triangle BCD\)with AA similararity.


We can find the side lengths of the triangles relative to each other, where \(\displaystyle x\) is the length of BC and \(\displaystyle r\) is the ratio of \(\displaystyle \triangle BCD\) to \(\displaystyle \triangle ABC\).




using the angle bisector theorem, we can solve for the ratio.


\(\displaystyle \frac{x\div r}{x\div r - x \cdot r} = \frac{x}{x \cdot r}\)


\(\displaystyle \frac{1}{1 - r^2} = \frac{1}{r}\)


\(\displaystyle 1- r^2 -r = 0\)


\(\displaystyle r=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-1)}}{2\cdot1}\)


\(\displaystyle r=\frac{-1\pm\sqrt{5}}{2}\)


because \(\displaystyle D\)is on \(\displaystyle \overline AC\), \(\displaystyle x\cdot r < x\div r, so -1>r>1\), meaning \(\displaystyle r\neq\frac{-1-\sqrt{5}}{2}\)because \(\displaystyle \frac{-1-\sqrt{5}}{2} < -1\).


plugging in the value of \(\displaystyle r\)into AD, we get

\(\displaystyle AD = x\div \frac{-1+\sqrt{5}}{2} -x \cdot \frac{-1+\sqrt{5}}{2}\)

\(\displaystyle AD = \frac{x\cdot(\sqrt{5}+1)}{2} - \frac{-x+\sqrt{5}x}{2}\)

\(\displaystyle AD = \frac{\sqrt{5}x+x+x-\sqrt{5}x}{2}\)

\(\displaystyle AD = \frac{2x}{2}\)

\(\displaystyle AD = x\)


Which means \(\displaystyle \boxed{BC = BD = AD}\).


Part b


We set \(\displaystyle y\)equal to \(\displaystyle x\cdot r\), our old value of \(\displaystyle CD\). This sets \(\displaystyle r\)equal to \(\displaystyle \frac{y}{x}\). We can define \(\displaystyle AB\)and \(\displaystyle AC\)both as \(\displaystyle x \div \frac{y}{x} = \frac{x^2}{y}\), or as \(\displaystyle x+y\).

So, \(\displaystyle \boxed{\frac{x^2}{y}=x+y}\).


Part c


We write the equation from above but using \(\displaystyle r\) instead of \(\displaystyle \frac{y}{x}.\)

\(\displaystyle x \div r = x+y.\)

\(\displaystyle x = (x+y)\cdot r.\)

\(\displaystyle \frac{x}{x+y}=r.\)

\(\displaystyle \boxed{r=\frac{x}{x+y}}.\)



I'm stuck on how to do part d.

I tried dropping an altitude from C to AD but I don't really know where to go from there
 
Last edited:

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
11,107
Part a can be done more quickly, just finding isosceles triangles. Most of the work you did there looks like the work for b and c and beyond.

Since you've already found the value of r, use it with the law of cosines and/or the law of sines with one or another of the triangles in the picture to answer d. Or, you could drop a perpendicular from A to BC, and from D to AB.
 
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