Maths help please..

[Ted]

That was my thought.
The ‘right angle’ looks to be obtuse (hence my comment!), and they are normally identified by a square iirc (it WAS over 50 years ago)

 

[Ted]

Double post

 

[Deleted member 65149]

The ratio of a triangle that will give a right angle is 3:4:5 there is no indication of angles so assuming it's a right angle at B I would guess BE would be 15cm though the 26cm should be 25cm.

Going back to my trigonmetry
Perpendicular/ Hypotenuse/ Sine
Base/ Hypotenuse/ Cosine
Perpendicular/Base/Tangent

The way I can still recall these are three little dittys from my Maths teacher, thank you G S King I at least learnt something.

Public House Sign
Cost of Bread High
Tan the Poor Boy

60 years ago I had no problem remembering the trig calculations. Then our maths teacher gave us some helpful hints that I found I needed from then on:
Sin = Passport to Hell
Cos(h) = Bash on Head
Tan = Pole on Bottom

(Of course the second one not to be confused with the hyperbolic cosine function, cosh.)

PS. Maths formed part of my studies up to MSc level but I had no easy solution to the original question. Well done mini Ant)

 

[baxlin]

There's no specification that there is a right angle involved.

The implication is that ABE and ACD are the same angle.

So that means BE is just going to be proportional to CD by the same ratio as AB is to AC.

And you can work out ED the same way - but it's not dependent on the assumption about the angle.

The diagram does show parallel line marks on BE and CD, so in my book this (^^^^) and post #3 are correct.

Easy when you know how!!

 

[Petrol Pete]

I wasn't in class that day.

I wasn't in class that whole week !

 

[Benzowner]

It was a sort of trick question, it is not about triangles, it is about ratios and the fact that the B-E and C-D are parallel. If they weren't, it could not be calculated with the information provided.

 

[Gollom]

Well I did not even understand the question...

 

[Darrell]

Well I did not even understand the question...

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