In right triangle ABC, tanA=0.4. In right triangle DEF, tanD=0.4 △ABC must be similar to △DEF.

a.) true
b.) false

Question

In right triangle ABC, tanA=0.4. In right triangle DEF, tanD=0.4 △ABC must be similar to △DEF.

a.) true
b.) false

anonymous · Answer

If both triangles have the same tan, does that mean that they are similar?

ankit042 · Answer

are the two triangle right angled?

ankit042 · Answer

I am not sure if this is 100% correct. tan for an angle in a triangle is generally defined for right angled triangles  so we can use SAS property to prove triangles are similar.

mathstudent55 · Answer

In $ \color{red}{\text{right}} $ triangle ABC, tanA=0.4. In $ \color{red}{\text{right}} $ triangle DEF, tanD=0.4 △ABC must be similar to △DEF.

a.) true
b.) false

mathstudent55 · Answer

|dw:1374724343363:dw|

anonymous · Answer

Okie dokie! So this would make it true, correct?

anonymous · Answer

Thank you for the very nice visual! Excellent help!

ankit042 · Answer

@mathstudent55 thanks for help! i forgot to look the right angel in the question

mathstudent55 · Answer

From the figure, we have 

$ \dfrac{BC}{AC} = \dfrac{EF}{DF} $

Sqaring both sides:

$ \dfrac{(BC)^2}{(AC)^2} = \dfrac{(EF)^2}{(DF)^2} $

From a property of proportions, we have:

$ \dfrac{(BC)^2 + (AC)^2}{(AC)^2} = \dfrac{(EF)^2 + (DF)^2}{(DF)^2} $



From the right triangles, we have:

$ (AC)^2 + (BC)^2 = (AB)^2 $

$ (DF)^2 + (EF)^2 = (DE)^2 $

By substitution, we get:

$ \dfrac{(AB)^2}{(AC)^2} = \dfrac{(DE)^2}{(DF)^2} $

By a property of proportions, we get:

$ \dfrac{(AC)^2}{(DF)^2} = \dfrac{(AB)^2}{(DE)^2} $

Take square roots:

$ \dfrac{AC}{DF} = \dfrac{AB}{DE} $

$ \dfrac{BC}{AC} = \dfrac{EF}{DF} $

$ \dfrac{BC}{EF} = \dfrac{AC}{DF} $

$ \dfrac{BC}{EF} = \dfrac{AC}{DF} = \dfrac{AB}{DE}$

Since the lengths of all sides of the one triangle are proportional to the lengths of all corresponding sides of the other triangle, the triangles are similar.

mathstudent55 · Answer

Yes, it's true.

anonymous · Answer

Thank you for teaching me how to thoroughly figure this out!! 

You're awesome! =)

mathstudent55 · Answer

You're welcome.

mathstudent55 · Answer

I think @ankit042 got it right way above with using SAS Similarity.

anonymous · Answer

Yes, good job @ankit042!

ankit042 · Answer

Thanks and you are welcome. @catsarecool1995  you know how to use SAS or need explination?

anonymous · Answer

Yes thank you very much!!♥