# Thread: trigo problem no. 12

1. ## trigo problem no. 12

p202 q20 re.ex.9
don't know how to do [b][ii]---show that ABC is equilaqteral. thanks

2. Hello, afeasfaerw23231233!

I've made one observation . . . $\Delta ODE$ is equilateral.
. . But I haven't gone any further.

We have these areas: . $\begin{Bmatrix}\Delta CED & = & \frac{1}{2}ab\cos^2\theta\sin\theta \\ \text{quad }ADEB & = & \frac{1}{2}ab\sin^3\!\theta \end{Bmatrix}$

We are given the ratio of these areas: . $\frac{\Delta CED }{\text{quad }ADEB } \;=\;\frac{\frac{1}{2}ab\cos^2\!\theta\sin\theta}{ \frac{1}{2}ab\sin^3\!\theta} \;=\;\frac{1}{3}$

. . which simplifies to: . $\frac{\cos^2\!\theta}{\sin^2\!\theta} \:=\:\frac{1}{3}\quad\Rightarrow\quad\tan^2\!\thet a \:=\:3\quad\Rightarrow\quad \tan\theta \:=\:\sqrt{3}$

Hence: . $\theta \:=\:60^o$

Since $\Delta ODE$ is isosceles $(OD = OE = \text{radius}),$
. . then: . $\angle DEO = 60^o$ . . . Hence: . $\angle DOE = 60^o$

Therefore: . $\Delta ODE$ is equilateral.

3. sorry, i made a calculating mistake, theta is not = 45 degree!
but i still cannot prove triangle ABC is equilateral