In right triangle ABC, we have angle BAC = 90° and D is the midpoint of line AC. If AB = 7 and BC = 25, then what is tan of angle BDC?

 Jun 11, 2017

BTW I need this quick! :)

 Jun 11, 2017

Use Pythagoras theorem:


\(CD=DA=12\)(mid point)

Use Pythagoras theorem again:


Find the sin of angle BCD:

\(\sin \angle BCD =\sin \angle BCA = \dfrac{7}{25}\)(angle BCD and angle BCA refers to the same angle)

Use law of sines on triangle BCD:

\(\dfrac{BD}{\sin\angle BCD}=\dfrac{BC}{\sin\angle BDC}\\ \dfrac{\sqrt{193}}{\frac{7}{25}}=\dfrac{25}{\sin\angle BDC}\\ \sin\angle BDC = \dfrac{25\times \frac{7}{25}}{\sqrt{193}}=\dfrac{7}{\sqrt{193}}\)

Construct a right-angled triangle with

legs = 7, -12(<-- negative because the angle is in the 2nd quadrant), hypotenuse=sqrt193

\(\tan \angle BDC =-\dfrac{7}{12}\)

 Jun 11, 2017

Quicker way:

angle BDC = 90 deg + angle DBA

\(\boxed{\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a \cos b+\cos a \sin b}{\cos a \cos b - \sin a \sin b}=\dfrac{\tan a + \tan b}{1-\tan a \tan b}}\)

But because tan 90 deg does not have a meaning, we use the old definition:

\(\tan 90^{\circ}=\dfrac{1}{0}\)


\(\dfrac{1}{\tan 90^{\circ}}=0\)

\(\tan \angle BDC \\= \tan (90^{\circ}+\angle DBA)\\ =\dfrac{\tan 90^{\circ}+\tan{\angle DBA}}{1-\tan90^{\circ}\tan\angle DBA}\\ =\dfrac{1+\frac{\tan \angle DBA}{\tan 90^{\circ}}}{\dfrac{1}{\tan 90^{\circ}}-\tan \angle DBA}\\ =\dfrac{1+0\tan\angle DBA}{0-\tan\angle DBA}\\ =-\dfrac{1}{\tan\angle DBA}\)

AC = 24 using Pyth. thm. <-- I am lazy in typing

AD = 24/2 = 12

So tan angle DBA = 12/7

So tan angle BDC = -1/tan angle DBA = -7/12 <-- done :)

 Jun 11, 2017


Since BDA and BDC are supplemental, tan( BDA)   =  - tan (BDC) →  - tan (BDA)  = tan (BDC)


Using right angle trigonometry,  tan (BDA)  =  7/12


So  -tan(BDA)  =  -7/12  = tan (BDC)



cool cool cool

 Jun 11, 2017

8 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.