Answer the following Trigo question (Re-post)?

I think you meant minimum value of angle CAD possible. If you really meant maximum value, then the answer is simply 90 degrees like kritan pointed out.

Solve it like this

http://answers.yahoo.com/question/index;_ylt=Amaqy...

Let w be the angle CAD

You have

(x/29 + tanw)/(1 - x/29 * tanw) = 2x/29

cross multiply and simplify in terms of descending powers of x and solve for x using quadratic formula

x = -(29cotw/4)(-1 +/- √(1 - 8tan²w))

For x to be "can't be found", the discriminant is < 0

1 - 8tan²w < 0

==> tan²w > 1/8

==> tanw > √(1/8)

==> w > 19.47122063 degrees, take positive value only since w is an acute angle

So the minimum angle is arctan(√(1/8)) = 19.4712206344906913692459993399624359630 degrees as Scythian1950 points out

Note that for your previous question, I mistakenly solve for tanw = 1/8, which gives 7.125016347 degrees.

Because this is a right triangle, Angle CAB is 45 degrees.

First, start by getting the length of Line AD. Since we know that Angle CAB is 45 degrees, Angle DAB must be 45-20=25 degrees. To get the length of Line AD, use: COS=Adjacent/Hypotenuse, which is COS 25 degrees= 29/Line AD. COS 25=.906. We get .906AD=29; divide out the .906 and you get a measurement for Line AD as 29/.906= 32 cm.

Now that we have the length of line AD, we can determine the length of line CD by using:

Tan=Opposite/Adjacent. Tan 20 degrees=Line CD/32. We proceed to get .364=Line CD/32. Multiply both sides by 32 to determine the length of Line CD and it measures .364*32= 11.65 cm.

Angle BAD = a , AD is the median

--->tan a = x/29 , tan(a+ 20) = 2x/29

---> 2tan20 x^2 - 29x + 29^2 tan20 = 0 ---> no solution for x

let angle DAC = t

2tan t x^2 - 29x + 29^2 tan t = 0

dscriminant = 29^2 (1 - 8 tan^2 t) >or=0 ----> tan t < 1/(2sqrt2) ----> t max = 19.47 degrees

My idea is this :

in triangle ACD, x/sin20=AD/sinACD (law of sines) and in triangle ADB, x/sint = AD/sin90 where t is angleDAB and angle CBA = 90

x = AD*sin20/sinACD = AD*sint/sin90 where we'll replace angle ACD as [90 -(20+t)]

and sin[90 -(20+t)] = cos(20+t)

On solving we get,

sint = (2 - cos^2(t))/(cost*cot20)

for angle CAD to be max, t has to be min and so sint will be min

so differentiating w.r.t t and further solving

we get cos^2(t)*[cot20 + 1+ cot^2(20)] = 2cosec^2(20)

and so t can be found out.Is this valid?

Sir, is the angle CAD still 20 deg if AD is meadian and not angle bisector?

gôhpihán:

Minimum value for angle CAD can be very close to zero when side BC is close to zero, so as Mr. Madhukar has asked search should be for maximum value for CAD.

Maximum value of angle BAC can be 90 but maximum value for angle CAD is arctan(√(1/8)) as you have rightly pointed out.

Just to note that this maximum value is independent of length of AB so 29 is irrelevant.

Refer to the problem

http://www.imagebam.com/image/55f9ae1825%E2%80%A6

(a) if AD is the bisector of angle CAB,

x = CD = CB -- DB = 29[tan40 -- tan20] = 29(0.8391 -- 0.3640) cm = 13.8 ANSWER

90