tan(angle XYZ) = XZ/XY

given slope of line ZY = -5 which is equal to -1 * tan(angle XYZ)

Therefore tan(angle XYZ) = 5

5= XZ/2

XZ = 10

Can u solve it another way ?

If you use equation of slope \frac{y_2-y_1}{x_2-x_1} and right triangle equation H^2=P^2+B^2, you’ll get an answer in the lines of 10.xxx

I’m going to name angles as <X, <Y, <Z corresponding to the vertices. For line `YZ`

, you know the slope is `-5`

. Also this slope equal to `-tan(<Y)`

( trigo property :` tan(180* - <Y) = -tan(<Y)`

). Also we know `tan(<Y) = ZX/XY`

i.e. `-(slope of ZY) = ZX/XY`

. Solving this equation gives `ZX = 5`

. Btw, you don’t need that trigo property, as when you solve with `tan(<Y) = -5`

, you’ll get `ZX = 5`

. Since length can’t be negative, it must be `5`

then.

@HoldMyBeer it worked for me, thank you

Got exactly 10 after aligning the triangle to 1st quadrant.

This is much quicker and easier. Thanks.