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1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\overline{XY}$ at $Z.$ Find $YZ.$

 

[asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label("$W$", A, N); label("$X$", B, SW); label("$Y$", C, SE); label("$Z$", D, SW); [/asy]

 

2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below.

 

[asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label("$A$", A, NE); label("$P$", P, SW); label("$Q$", Q, NE); label("$R$", R, E); label("$T$", T, NW); label("$U$", U, S); [/asy]

 

If $\angle QTA = 41^\circ$ and $\angle RUA = 63^\circ,$ then find $\angle QPR,$ in degrees.

 Feb 13, 2020
 #1
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+1

1) By similar triangles, YZ = 1/2*14^2/8 = 49/4.

 

2) Angle PQR = 90 - (41 + 63)/2 = 38 degrees.

 Feb 13, 2020
 #2
avatar+108727 
0

Yes, seeing lazily posted nonsense like this on the forum is enough to make me sick too.

Though MaxWong appreciated your coding.   Maybe if I was learning coding, which I would like to one day, I would appreciate it too. 

But as a maths question, it is close to useless.

 Feb 13, 2020
edited by Melody  Feb 13, 2020
 #3
avatar+302 
+1

I think this is in LaTeX. I can understand the first two lines. But on the second and 4th paragraph it's all giberrish.

 Feb 13, 2020
 #4
avatar+108727 
0

It is not Latex. Maybe it is C++  but I am not sure.

Melody  Feb 13, 2020
 #5
avatar+531 
+2

Something's fishy with the question #1  cheeky

The pic shows  how is the tangent line drawn. 

 

 Feb 13, 2020

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