On-line waiting for math problems in grade three

( 1)

Given ctgA=4/3, we assume that AC=4a and BC=3a.

From Pythagorean Theorem: (4a)? +(3a)？ = 10? = 100

= = = & gt25a？ = 100

= = = & gta=2

So, AC=8, BC=6.

Known ∠ PBC = ∠ A.

So ctg∠PBC=BC/PC=4/3.

= = = & gt6/ individual =4/3

= = = & gtPC=9/2

So AP=AC-PC=8-(9/2)=7/2.

(2) As shown in the figure, let the circle O be tangent to D with AC and tangent to E with AB to connect o D and O E..

Then, OD⊥AC, OE⊥AB.

And od = OE = y.

Then, s △ OPA = (1/2) pa * od = (1/2) xy; s△OAB =( 1/2)AB * OE =( 1/2)* 10 * y = 5y

Therefore, S△BPA=( 1/2)xy+5y.

And s △ BPA = (1/2) pa * BC = (1/2) * x * 6 = 3x.

So: (1/2)xy+5y=3x.

= = = & gtxy+ 10y=6x

= = = & gt(x+ 10)y=6x

= = = & gty=6x/(x+ 10)

Because p is on the line segment AC, 0 < x < 8.