you have two similar RIGHT triangles, triangle TAB and triangle MCB.?

Question

angle A is the right angle in triangle TAB and angle C is the right angle in triangle MCB. (triangle 1) TA=X, AB=125 FT, TB=?, (triangle 2) BC=50FT, CM=30 FT, MB=?

how can i find x?

Answer 1

Assuming that the similarity exists in the way the triangles are named, it would mean that TA/MC = AB/CB = TB/MB. Substituting the values given, x/30 = 125/50 = TB/MB.
Solving for x, x=30*125/50 = 75 feet.
From triangle MCB, MB squared=BC squared plus MC squared as angle C is a right angle. i.e., MB squared = 2500+900 and so MB = sq. root of 3400 = 58.3 feet, to one decimal place.
Hence TB=MB*125/50 = 58.3*2.5 = 145.8 feet., to one decimal place.

Answer 2

Had to draw this one out.

Now use this as a ratio. Actually its a proportion to solve.

50ft is to 125ft as 30ft is to x.

50/125 = 30/x

50x = 3750

x = 75