Triangle RST is drawn on the coordinate plane with vertices at (5,−6) , (5,0) , and (7,2) . The triangle is then translated 2 units to the right and 4 units down to create triangle R′S′T′ . What is the sum of the angles of triangle R′S′T′ ?(1 point)
The sum of the angles is
°.
To find the sum of the angles of triangle R'S'T', we first need to find the lengths of the sides of triangle RST, and then use those lengths to find the angles.
The length of RS is given by the difference in x-coordinates of the two points: RS = 5 - 5 = 0.
The length of ST is given by the difference in y-coordinates of the two points: ST = 2 - 0 = 2.
The length of RT can be found using the distance formula: RT = sqrt((7 - 5)^2 + (2 - (-6))^2) = sqrt(2^2 + 8^2) = sqrt(4 + 64) = sqrt(68).
To find the angles, we can use the Law of Cosines. Let RST be angle x, SRT be angle y, and STR be angle z.
Using the Law of Cosines, we have:
RT^2 = RS^2 + ST^2 - 2(RS)(ST)cos(x)
68 = 0^2 + 2^2 - 2(0)(2)cos(x)
68 = 4 - 0
68 = 4
This is not a valid equation, so there is no angle x.
To find angle y, we can use the Law of Cosines again:
ST^2 = RS^2 + RT^2 - 2(RS)(RT)cos(y)
2^2 = 0^2 + 68 - 2(0)(sqrt(68))cos(y)
4 = 68 - 0
4 = 68
This is also not a valid equation, so there is no angle y.
Finally, to find angle z, we can use the Law of Cosines:
RS^2 = ST^2 + RT^2 - 2(ST)(RT)cos(z)
0^2 = 2^2 + 68 - 2(2)(sqrt(68))cos(z)
0 = 4 - 4(sqrt(68))cos(z)
Since the right side of the equation is zero, we must have:
0 = sqrt(68)cos(z)
Since cosine is never negative, we can conclude that cos(z) = 0.
Therefore, angle z must be a right angle (90°).
Since the sum of the angles in a triangle is always 180°, we have:
Angle x + Angle y + Angle z = 180°
0° + 0° + 90° = 90°
So, the sum of the angles of triangle R'S'T' is 90°.