finding the cosine

homeschool girl

Junior Member
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
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.