1
$\begingroup$

In triangle ABC the angle at A is 60 and the inscribed circle touches AB at the point D. If AD = 5 and DB = 3, find the length of BC. enter image description here

I know BD = BE since they are tangents and AD = AF but this is the furthest I've reached into solving this problem but I can't think of anything else to do.

$\endgroup$
2
$\begingroup$

Let $O$ be the center of the inscribed circle. Then $\Delta ADO$ forms a 30-90-60 $\Delta$. Hence you get the radius $r$ of the inscribed circle. Use this in $\Delta OBE$ to get angle $B$ and hence angle $C$ and hence $EC$.

$\endgroup$
  • $\begingroup$ how do I get the radius $r$ of the circle? if you are saying that $ΔADO$ forms a 30-90-60 then how do I proceed from there? $\endgroup$ – Aspiring Mathlete Mar 9 '14 at 16:37
  • $\begingroup$ $AD$ is given and you know $\angle OAD$. Using trigonometric formula for $\tan$ you can calculate $r$. $\endgroup$ – Sandeep Thilakan Mar 10 '14 at 4:58
  • $\begingroup$ But I do not know the value for AO nor do I know the value for OD. Also, I am not allowed to use a calculator. $\endgroup$ – Aspiring Mathlete Mar 10 '14 at 15:43
  • $\begingroup$ $\tan \angle OAD = \tan 30 = \frac{OD}{AD}$ Therefore $r = \frac{5}{\sqrt{3}}$ $\endgroup$ – Sandeep Thilakan Mar 11 '14 at 5:43
  • $\begingroup$ OK. I know now that $OE$ = $r$. Now how do I proceed? I cannot get the value of angle B without a calculator and I am not allowed to use one. Also, the memorandum says that that BC = 13, but I do not know how to get that answer. Please can you show me how you get the value of BC? $\endgroup$ – Aspiring Mathlete Mar 11 '14 at 15:47
0
$\begingroup$

there is property of circle construction :Inscribe a Circle in a Triangle

Steps:

Bisect one of the angles
Bisect another angle
Where they cross is the center of the inscribed circle

now if you bisect one angle ,namely angle $A$,but this also would be right triangle,because The angle between the tangent and radius is 90°,example enter image description here

and you can easily proceed from here

$\endgroup$
0
$\begingroup$

The tangents CF and CE are also equal - call their value $x$, then apply the cosine rule. The extent of the trigonometry required is knowing that $\cos(60) = \frac{1}{2}$.

$\endgroup$

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.