Easy Question (1 Viewer)

[kurt.physics]

Easy but i cant get it, it is as follows

In the diagram, PQRS is a square and STR is an equilateral triangle in the same plane. Find angle PTQ.

A) 16* B) 22*30' C) 30* D) 36* E) 40*

(* is degrees)

The diagram is attached

 

[conics2008]

its C..

30*

you said its a square therefore that triangle you speak of has the same length.. hence draw a triangle

with sides 2root of 2+root3 one side and the other crosspoding side and the opposite side length 2

use cos x= 2rootof 2+roo3 ^2 + 2root 2+root3 ^2 -4 / 2 * 2root 2+ root 3 ^2 = cos x = root3/2 = 30*

 

[shaon0]

its C..

30*

Wat year work is this? its obviously 30 degrees

 

[Mark576]

Simplest way to do it is to notice that ∠TRQ = 150 degrees and since triangle TRQ is isoceles (TR = RQ) ∠RTQ = ∠STP (RTQ ||| STP)(obv. they're also congruent) = 15 degrees. The result follows. You could also set the length of the square to 1 unit on an x-y cartesian co-ordinate system. Then find the gradients of TQ and PT and use the formula for the angle between two lines to find the answer, 30 degrees, (C).

 

[conics2008]

I dont know man, it just all connected... find angle PTQ not STP

 

[conics2008]

Wat year work is this? its obviously 60 degrees

it cant be 60* ...

you just assume the other triangle ptq is an equally triangle

 

[shaon0]

My bad did i write 60 degrees meant to be 30 degrees sorry.

 

[conics2008]

hey kurt where do you find these questions...

which books do you use for maths...

i only have 4unit cambridge and fitzpatrick.. and 3unit year 11 cambridge...

 

[ronnknee]

Does this really incorporate 4U material?

 

[conics2008]

ronnknee its for masturbation purpose!

 

[kurt.physics]

its C..

30*

you said its a square therefore that triangle you speak of has the same length.. hence draw a triangle

with sides 2root of 2+root3 one side and the other crosspoding side and the opposite side length 2

use cos x= 2rootof 2+roo3 ^2 + 2root 2+root3 ^2 -4 / 2 * 2root 2+ root 3 ^2 = cos x = root3/2 = 30*

Im not sure this is right, no sides were given so i cant see how you can say the side of the triangle is '2root of 2+root3' and deduce it from there. The side of the square is the same as any one of the sides of the triangle, its an equilateral triangle and one side of the triangle meets with the square...hence all sides are equal in the triangle and the square combined.

This question came from the Australian Mathematics Competition Intermediate Division (Years 9-10).

I doubt that you need trigonometry for achieveing the solution as the AMT (Australian Mathematics Trust) do not usually use trig in the tests because you cannot use calculators (even if you know the angles and their corresponding exact sides of by heart)

 

[kurt.physics]

Simplest way to do it is to notice that *TRQ = 150 degrees and since triangle TRQ is isoceles (TR = RQ) *RTQ = *STP (RTQ ||| STP)(obv. they're also congruent) = 15 degrees.

OMG, this is a brilliant solution, Triangle TRQ is issoseles and therefore angle TQR = angle QTR so the angle of TQR is 15

Well done

 

[conics2008]

KURT in maths sometimes you can assume the length of the sides... plus there are hundred of methods on solving that question, for me thats how i saw it ..

i can give length to those sides and find it.. you can plot them on the x-y axis and find it from there....

thats how i found it....

 

[kurt.physics]

plus there are hundred of methods on solving that question, for me thats how i saw it ..

i can give length to those sides and find it.. you can plot them on the x-y axis and find it from there....

thats how i found it....

I know that many problems have hundreds of ways of solving them but Mark576's way is probably the most easiest, even though your method is more elegant, its faster (especially in the AMC where its more of a race of time than what you know).

 

[ronnknee]

ronnknee its for masturbation purpose!

Right