Answer:
Anika is correct
Step-by-step explanation:
* Lets revise the rules of reflection and rotation
- If point (x , y) reflected across the x-axis then its image is (x , -y)
- If point (x , y) rotated about the origin by angle 270° clock wise then
its image is (-y , x)
* Lets check the vertices of the two triangles
- The vertices of Δ ABC are:
# A (-7 , 2)
# B (-3 , 6)
# C (-2 , 1)
- The vertices of Δ PRQ
# P (-7 , -2)
# R (-3 , -6)
# Q (-2 , -1)
* By comparing between the vertices of the two triangles
∵ Each y-coordinates of Δ PRQ has opposite sign of each
y-coordinates of Δ ABC
∵ All x-coordinates of Δ PRQ are the same with x-coordinates of
Δ ABC
- That means Δ PRQ is the image of Δ ABC after reflection across
the x-axis
∵ Reflection doesn't change the shape and the size of the figure
∴ Δ ABC and Δ PRQ have same size (equal sides and equal angles)
∴ Δ ABC is congruent to Δ PRQ
* Anika is correct
Answer:
Mya only used the dimensions from the scale drawing of a rectangle
Step-by-step explanation:
A
J = 7 × 6
j = 42
j is 7 times greater than 6, so you have to put j equal to 7 times the number 6 :)
Cos 52 = 0.6157
sin 42 = cos 48 = 2/3 (because 42 + 38 = 90 - complementary angles)
if tan 20,5 = 3/8 then tan 69.5 wil = 1 / 3/8 = 8/3