In the figure, four line segments A B , C D , E F , G H intersect at a single point.
Find the value (in degrees) of the angle sum x + y + w + z .
Note
:
∠
E
F
H
=
5
5
∘
,
∠
A
C
D
=
3
7
∘
, and
∠
G
H
F
=
8
5
∘
.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
simply good....
There is nothing to stop AB from being on the same line as EF or the other way so DC are on the same line as GH
Too easy question..... Just apply " Sum of all angles of a triangle equals 180° " & " Sum of all angles on a point is 360° " . :p
2 (180 - 37 - 90) = 106
2 (180 - 85 - 55) = 80
360 - 106 - 80 = 174
sum of all angles around a point is 360*
Os triângulos ACDB, são iguais, então os ângulos correspondentes vão ser os mesmos, assim como nos triângulos EGO,OFH( O >>PONTO DE ORIGEM DO SISTEMA), sabendo que a soma dos ângulos internos de um triângulo é 180º: triângulo ACO,ODB >> 180-(37+90)=53º
triângulo EGO,OHF>>180-(85+55)=40º
Sabendo que uma volta inteira é igual a 360º
LOGO, x+y+w+z= 360- (53 X 2)+(40 X 2): 360-106-80=360-186=174º
Resposta : 174º
Could'nt agree with u more...
As you can see every triangles and its opposite triangles are the same so you just find the missing angle and then add them all and then deduct it from 360° .. 360-186= 174..
It's 174.
First, I added the right angle plus 37 then subtracted that from 180. Then did it too the BD yriangle becasue they're vertical angles. Then I added 85 + 55 which I subtracted from 180. Then I did it too EG triangle because of vertical angles. Then subtracted by 180. Then all 4 triangles equaled 186. Then subtracted from 360, which equalled 174.
suppose think point O
then O1+y+O2+z=180 O3+x+O4+w=180 then y+z= 87, x+w=87 as sum of angles of triangle is 180 we can find them. so x+y+w+z = 87+87 = 174
I did it this way tho.LoL
180-(90+37)=53
180-(55+85)=40
2(53+40)=186
360-186=174
I did the hard way lol
180-(90+37)=53
180-(85+55)=40
2(53+40)=186
360-186=174
Write a solution. 180-(85+55)=40 180-(90+37)=53 360-(40+40+53+53)=x+ y+ w+ z=174
Taking the straight line AB z+y=180-(180-85-55)-(180-90-37)=180-40-53=87 As per alternate angles x=z and w=y so similarly x+w=87 Therefore, x+y+w+z=87+87=174
If we take the triangle that has angles H and F with angles of 85 and 55 degrees respectively. Since, the angles in a triangle add up to 180 degrees. We can take
180 - ( H + F ) = 180 - ( 85 + 55 ) = 180 - 140 = 40
If we take the other triangle with angles A and C with 90 and 37 degree angles respectively. From before we know that the three angles of triangle adds up to 180.
180 - ( A + C ) = 180 - ( 90 + 37 ) = 180 - 127 = 53
We also know the angle that is on the other side of the intersection point is the same angle. We also know that a sum of all angles around a point is 360 degrees . We can progress with
360 - 2 ( 40 + 53 ) = 360 - 186 = 174
So
x + y + w + z = 174
Let's say the centre or intersection point of the 4 lines is O. Angle AOC = angle DOB, which is (180-90-37= 53)°; just as angle HOF = angle EOG, which is (180-85-55=40)°.
Following these angles, it is possible to determine angle w+y+x+z. The question does not ask for each individual angle, therefore (I use < as the angle sign) :
<AOC + <y +<HOF +<z +<DOB +<w +<GOE +<x =360°
53°+<y+40°+<z+53°+<w+40°+<x=360°
<(w+x+y+z)=360°-(53+40+53+40)°
<(w+x+y+z)=(360-186)°
<(w+x+y+z)=174°
Problem Loading...
Note Loading...
Set Loading...
2 (180 - 37 - 90) = 106
2 (180 - 85 - 55) = 80
360 - 106 - 80 = 174