Answer:
there will be 4 cornered shapes 1 rectangle that is almost a square and the shape I Forgot TvT
Let stadium 1 be the one on the left and stadium 2 the one on the right.
Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).
For the next part we need to use the trigonometric function of tangent.
As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)
Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)
Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)
Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)
Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m
Correcting the answer to whole number gives you 1754 m which is the option C.
27 is 144% of 18.75
mk, hope I helped
Answer:
x=6
y=3
Step-by-step explanation:
y(side)= x
x(side)=2x
3rd side (3
) = x
x=3;
x= 2(3)=6
y=(3)=3
Answer:
1) AD=BC(corresponding parts of congruent triangles)
2)The value of x and y are 65 ° and 77.5° respectively
Step-by-step explanation:
1)
Given : AD||BC
AC bisects BD
So, AE=EC and BE=ED
We need to prove AD = BC
In ΔAED and ΔBEC
AE=EC (Given)
( Vertically opposite angles)
BE=ED (Given)
So, ΔAED ≅ ΔBEC (By SAS)
So, AD=BC(corresponding parts of congruent triangles)
Hence Proved
2)
Refer the attached figure
In ΔDBC
BC=DC (Given)
So,
(Opposite angles of equal sides are equal)
So,
So,
(Angle sum property)
x+x+50=180
2x+50=180
2x=130
x=65
So,
Now,
So,
In ΔABD
AB = BD (Given)
So,
(Opposite angles of equal sides are equal)
So,
So,
(Angle Sum property)
y+y+25=180
2y=180-25
2y=155
y=77.5
So, The value of x and y are 65 ° and 77.5° respectively
