Which triangle is similar to △ABC if sin(A) = One-fourth, cos(A) = StartFraction StartRoot 15 EndRoot Over 4 EndFraction, and tan(A) = StartFraction 1 Over StartRoot 15 EndRoot EndFraction?

ΔXYZ

Step-by-step explanation:

Trigonometric identities are equations which are true for right angled triangles. They are:

sin(A) = opposite / hypotenuse, cos(A) = adjacent / hypotenuse, tan(A) = opposite / adjacent.

Given triangle ABC with ∠A with side BC opposite to the angle, ∠B with side AC opposite to the angle, ∠C with side AB opposite to the angle.

sin(A) = opposite / hypotenuse = BC / AC = 1/4

BC = 1, AC = 4

cos(A) = adjacent / hypotenuse = AB / AC = √15 / 4

AB = √15

tan(A) = opposite / adjacent = BC / AB = 1 / √15

Two triangles are said to be similar if the ratio of their corresponding sides are in the same proportion.

ΔXYZ and ΔABC are similar because the ratio of their corresponding sides are in the same proportion.

For adjacent side; XZ / AB = 6√15 / √15 = 6

For opposite side; YZ / BC = 6/1 = 6

For hypotenuse side; XY / AC = 24 / 4 = 6

D. ΔXYZ

Step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.