0
$\begingroup$

enter image description here

I want to determine the lenght of $x$ by making equations for each triangle ACD and ADB.

Let's recall $\angle DAB = \alpha $, $\angle DBA = \beta $ in triangle ADB

$$\alpha + \beta + 90 = 180$$

$$\alpha + \beta =180 \tag{1}$$

In isosceles triangle ACD, $\angle CAD = \angle CDA = \alpha$ and let $\angle ACD = \theta $

$$2\alpha + \theta = 180 \tag{2}$$

From two equations, we can conclude that $\alpha = 2\beta$. However, this does not actually help me at all. What am I missing?

Regards

$\endgroup$
1
$\begingroup$

@Enzo Call $t=AC=CD=DB$. From Pythagorean Theorem you have $$ 10^2=t^2+x^2 $$ From the law of cosines in triangle $ACD$ you get $$ 2t^2-2t^2\cos\theta=x^2 $$ But $\cos\theta=\cos(180-2\alpha)=-\cos(2\alpha)=1-2\cos^2\alpha=1-2(t/10)^2$. If you plug this value in the second displayed equation above you get a system for $x$ and $t$ that reduces to a quadratic equation for either $t^2$ or $x^2$

$\endgroup$
1
$\begingroup$

You can conclude by observing that $\angle ADB$ is inscribed in a semicircle and is therefore right. This means that $\alpha + \beta = 90$, so $\alpha = 30$ and $\beta = 60$. It follows that $x = \frac{\sqrt{3}}{2}(10) = 5 \sqrt{3}$.

$\endgroup$
0
$\begingroup$

$$\alpha + \beta + 90 = 180, \, \beta= 2 \alpha,\,3 \alpha= 90, \, \alpha=30,\, \beta=60$$

Use Rule of Sines on right triangle $ \Delta ADB$ inside semi-circle $ACDB$

$$ \dfrac{x}{\sin 60}= \dfrac{10}{\sin 90} \rightarrow x= 10 \sqrt{3}/2 $$

$\endgroup$
  • $\begingroup$ There's a typo, I meant to write $\theta = 2\beta$. How can we conclude that it is also right for $ 2\alpha = \beta$? $\endgroup$ – Melz Jan 22 at 18:46
  • $\begingroup$ You should correct your question to be without typo. $\endgroup$ – Narasimham Jan 22 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.