Calculator, Q 17 - what it getting at


by thomas49th
Tags: calculator
thomas49th
thomas49th is offline
#1
Apr7-07, 07:46 AM
P: 656

What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
cristo
cristo is offline
#2
Apr7-07, 07:51 AM
Mentor
cristo's Avatar
P: 8,287
Quote Quote by thomas49th View Post
What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
Pythagoras?
thomas49th
thomas49th is offline
#3
Apr7-07, 07:55 AM
P: 656
[tex](x+8)(x+8) = x^{2} + (x+5)(x+5)[/tex]

[tex]x ^ {2} + 16x + 64 = x^4+10x+25[/tex]

take LHS from RHS

[tex]x^{2} - 6x - 39 = 0[/tex]

but how did you know to use pythagerous?

arildno
arildno is offline
#4
Apr7-07, 07:56 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016

Calculator, Q 17 - what it getting at


What relationships DO you know hold for right-angled triangles that might come in handy?
cristo
cristo is offline
#5
Apr7-07, 07:59 AM
Mentor
cristo's Avatar
P: 8,287
Quote Quote by thomas49th View Post
[tex](x+8)(x+8) = x^{2} + (x+5)(x+5)[/tex]

[tex]x ^ {2} + 16x + 64 = x^4+10x+25[/tex]

take LHS from RHS

[tex]x^{2} - 6x - 39 = 0[/tex]
Sorry, one correction to your second line.

[tex]x ^ {2} + 16x + 64 = 2x^2+10x+25[/tex]

but how did you know to use pythagerous?
Well, you did the hard part; spotting that it was a right angled triangle. Pythagoras' theorem holds for right angled triangles, and is a relationship relating the squares of the sides. Since the solution contains an x^2, this is quite a big hint as to what you should use.


Register to reply

Related Discussions
TI-89 Calculator Calculators 5
Calculator help Calculators 4
math ap Calculator Calculators 8
Do you really need a calculator? General Math 21
exp in calculator General Math 3