I have a right triangle 'ABC', with 'AB' as the hypotenuse. 'BC' and 'AC' are
3.5cm and 5.0cm respectively. I also have a midpoint 'm' betweem 'A' and
'C'. A line of length 'p' is drawn from m to the hypotenuse making an
angle (theta) with 'AB' facing 'BC'.

Alternatively, instead of having AC=3.5cm , you may make <A = 35 degrees.

The question is, find an expression for the length of 'p' expicitly in
terms of (theta). Preferable (theta) be expressed in radians.

A quick well worked out rsponse will generate a large tip.

observa-ga

Hi observer,
I'll solve this problem taking theta as if it were measured in
degrees. The conversion to radians will be done at the end. In order
to solve this problem, it's useful to think of the triangle's sides as
lines on a plane. This would be the graph:

 |y
 |
 |
 |      C
 |      /\_
 |     /   \_
 |    /      \_ m
 |   /       **\_
 |  /            \_
 | /               \_
 |/ B                \  A
 -------------------------------
 0                             x

Since we know that <A=35º and <B=90º (it's a right triangle), then
<C=55º. It's also easy to calculate the hypothenuse AC. Given that
<A=35º, then

AB = AC/cos(35º)
AB = 5 / 0.8192...
AB = 6.1039...

Now let's assume that theta is 90º (I'll do the conversion ot radians
later). We know that segment Am length is 2.5 (midpoint of AC, whose
length is 5). In this case, line 'p' is a vertical line from 'm' to
the x-axis (at this point, the x-axis represents AB). Let's call 'd'
the intersection between this vertical line and the x-axis:

 |y
 |
 |
 |      C
 |      /\_
 |     /   \_
 |    /      \_m
 |   /        |\_
 |  /        p|  \_
 | /          |    \_         (T=theta)
 |/ B        T|      \ A   
 -------------+-------+---------
 0            d                x

Since we know that <A=35º and Am=2.5, we can easily calculate the
length of segments dA and dm. So we have:

dm = 2.5*sin(35º)  (=1.433...)
dA = 2.5*cos(35º)  (=2.047...)

This gives the solution when theta is 90º: the length of 'p' is just
dm, or 2.5*sin(35º). Although the solution is clearly still
incomplete, the dm segment will be useful in finding the length of 'p'
for other cases of theta.

Let's analyze the two general cases of theta:


Case 1: Theta < 90º. This case looks like this:

 |y
 |
 |
 |      C
 |      /\_
 |     /   \_
 |    /      \_m
 |   /        |\_
 |  /          | \_
 | /            |  \_         (T=theta)
 |/ B           T|   \  A   
 -------------+---+-------------
 0            d   f            x

('f' is the point where 'p' intersects AB, or the x-axis)

Notice that we can form a right triangle dmf, and the length of 'p' is
just the length of segment mf, which is the hypothenuse of this
triangle. Since <f=T, then:

dm/mf = sin(T)
mf = 2.5*sin(35º)/sin(T)

Therefore, if T<90º, then the length of 'p' is 2.5*sin(35º)/sin(T)


Case 2: Theta > 90º. This case looks like this:

 |y
 |
 |
 |      C
 |      /\_
 |     /   \_
 |    /      \_m
 |   /        |\_
 |  /        |   \_
 | /        |      \_         (T=theta)
 |/ B     T|         \  A   
 ----------+--+----------------
 0         f  d                x

Again we have the right triangle dmf. However, this time, angle 'f' is
the complementary of T (notice that T is measuring the angle "outside"
triangle dmf. Therefore, <f=180º-T. Then we simply repeat the
reasoning:

dm/mf = sin (180º-T)
mf = 2.5*sin(35º)/sin(180º-T)

So, if theta is greater than 90º, then the length of 'p' is
2.5*sin(35º)/sin(180º-T).


So now we've done all the cases. The complete solution, with T
measured in degrees, with be like this:

                /
                | 2.5*sin(35º)/sin(T)        if T<90º
Length of 'p' = | 2.5*sin(35º)               if T=90º
                | 2.5*sin(35º)/sin(180º-T)   if T>90º
                \

Also, since sin(90º)=1, we can write this as:

                /
                | 2.5*sin(35º)/sin(T)        if T <= 90º
Length of 'p' = | 2.5*sin(35º)/sin(180º-T)   if T >  90º
                \

If we express T in terms of radians, since pi radians equal 180º, then we have:

                /
                | 2.5*sin((35/180)*pi)/sin(T)      if T <= (1/2)*pi
Length of 'p' = | 2.5*sin((35/180)*pi)/sin(pi-T)   if T >  (1/2)*pi
                \


I hope this helps! If you have any doubts regarding my answer, please
don't hesitate to request clarification before rating it.

Best wishes,
elmarto

WOW, comprehensive answer for just $2 :)

Here is another way.

Let d be the point where p touches AB.

a) If BC=3.5 and AC=5.

In right triangle ABC,
tan(angle A) = 3.5/5 = 0.7
So, angle A = arctan(0.7) = 34.992 = say, 35 degrees

In triangle Adm,
>>>angle Adm = (180 -theta) degrees.
>>>dm = p
>>>mA = 5/2 = 2.5 cm
>>>angle mAd = angle A = 35deg
By Law of Sines,
p/(sin A) = mA/(sin angle Adm)
So,
p/(sin 35deg) = 2.5/(sin (180-theta))
Multiply both sides by sin 35deg,
p = (2.5sin 35deg)/(sin 180-theta)
Since sin(180-theta) = sin(theta) also, then,
p = 2.5sin(35deg)/sin(theta)  <---***
p = 1.433941/sin(theta)  <---theta is in degrees.

b.) If angle A is 35 degrees.
It is the same as above.

----------

If theta is in radians, then change 35degrees into radians:
35 deg = (35deg)(pi rad /180deg) = 7pi/36 rad = 0.610865238 rad.
Then,
p = 2.5sin(0.610865238)/sin(theta)
p = 1.433941/sin(theta)  <---theta is in radians, 

Therefore, whether theta is in degrees or radians,
p = 1.433941/sin(theta)  <---answer.

That is for any value of theta. Be it be less, equal or greater than
90 degrees or pi/2 radians.
Be careful only on which mode, degrees or radians, you need to set
your calculator. Degrees when theta is in degrees; radians when theta
is in radians.

Hi: Although the answer appears to have been provided, let me give you
a completely "analytical" answer.  The triangle looks like this:

     A         I use the following notation:
     |\          <MAT = beta  (AC and BC are known; so we know beta);
     | \         <MTS = theta (this is angle that is given);
     |  \T          x = AM    (this is of course half of AC).
     | / \     So, in AMT, we have a triangle whose
     |/p  \         "base" length        = x;
    M-    --S       angles from the base = beta and (theta-beta).
     |      \  Therefore, we can simply solve this triangle to get
     |       \      p = length of MT
     |        \       = x*sin(beta)/sin(theta), where all angles are in rad.
     |         \    
     |__________\       In your problem statement, lengths of AC and BC are 
     C           B      given as 5.0cm and 3.5cm respectively.  Then,
                          x    = AC/2 = 2.5cm;
                          beta = tan^{-1}(BC/AC) = tan^{-1}(3.5/5.0) = 0.61 rad
So, simply plug in whatever value of theta, and you get the value of p.

For example, let us look at two "extreme" cases:
  CASE 1: When theta = pi/2 + beta (note: pi/2 rad = 90 deg), line MT
is parallel to BC and its length should be BC/2.  Indeed, the eqn
yields
          p = x*sin(beta) = BC/2.
  CASE 2: When theta = beta, line MT coincides with line MA and its
length should be x.  Indeed, the eqn yields
          p = x.

Hope this helps.

kprema

Regarding my comment added on 08/17, there is a minor type at the end:
In CASE 1, the eqn yields
  p = x*tan(beta) = BC/2.
But I am sure you already realized this.

kprema