 after the plane passes over the ground station, we get an angle between the measures a and b of 90 + 35 degrees. 
Awesome.
Now, let's build a table of values:
$$
a = 6\,\mathrm{km}\\
a' = 0 \quad \text{(duh)}\\
b = b'\cdot t = 4\cdot 6 = 24\\
b' = 4\,\mathrm{km}/\mathrm{min}\\
t' = 6min\\
c = \sqrt{a^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \approx 2.78316\\
c' =?
$$
Okay, let's differentiate the law of cosines, which is given to us in form of a hint:
$2cc' = 2aa' + 2bb' - \frac{du}{dt}$
$
u = 2ab\cos(\theta)$
$\frac{du}{dt} = \frac{ds}{dt} \cdot \cos(\theta) - \sin(\theta)\cdot(2ab)$
$\frac{ds}{dt} = 0\cdot ab + (a'b + b'a)$
And so now we rewind:
$\frac{du}{dt} = \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $
And so: 
$2cc' = 2aa' + 2bb' - \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $

Boom, boom. Let's substutute what we know
$2\left(2 \sqrt{15} \right)(c') = 2(6)(0) + 2(24)(4) - \left[ \left(0\cdot (6)(24) + \left((0)(24) +  (4)(6) \right) \right) \right] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$

Now what? Well, evaluate. Not sure
after the plane passes over the ground station, we draw an slant between the measures a and b of 90 + 35 academic degree. 
 Awesome. 
 immediately, let's build a board of value: 
 $ $ 
 a = 6\,\mathrm{km}\\ 
 a' = 0 \quad \text{(duh)}\\ 
 b = b'\cdot thymine = 4\cdot 6 = 24\\ 
 b' = 4\,\mathrm{km}/\mathrm{min}\\ 
 t' = 6min\\ 
 vitamin c = \sqrt{a^2 + b^2 } = \sqrt{36 + 24 } = 2\sqrt{15 } \approx 2.78316\\ 
 c' =? 
 $ $ 
 Okay, let's differentiate the law of cosines, which is contribute to us in form of a hint: 
 $ 2cc' = 2aa' + 2bb' - \frac{du}{dt}$ 
 $ 
 u = 2ab\cos(\theta)$ 
 $ \frac{du}{dt } = \frac{ds}{dt } \cdot \cos(\theta) - \sin(\theta)\cdot(2ab)$ 
 $ \frac{ds}{dt } = 0\cdot ab + (a'b + b'a)$ 
 And indeed now we rewind: 
 $ \frac{du}{dt } = \left [ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right ] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $ 
 And so: 
 $ 2cc' = 2aa' + 2bb' - \left [ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right ] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $ 

 Boom, boom. Let's substutute what we know 
 $ 2\left(2 \sqrt{15 } \right)(c') = 2(6)(0) + 2(24)(4) - \left [ \left(0\cdot (6)(24) + \left((0)(24) +   (4)(6) \right) \right) \right ] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$ 

 nowadays what? Well, evaluate. Not sure
 afher the plane passes ovev the ground stajiin, we jet an zngle begween the measures a and b oh 90 + 35 degrtvs. 
Awesome.
Now, let's bukld a tabpe of vaoues:
$$
e = 6\,\mathrm{km}\\
a' = 0 \qnzd \text{(duh)}\\
b = g'\gdot c = 4\cdot 6 = 24\\
b' = 4\,\matmrm{km}/\mathrm{kin}\\
t' = 6min\\
c = \sqst{x^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \approx 2.78316\\
c' =?
$$
Okay, let's dyfferenyiwte the law of cospnqs, wgpcm is given to us in form of a hiht:
$2cc' = 2ae' + 2bb' - \frac{du}{dt}$
$
i = 2ab\cos(\theta)$
$\frac{du}{dt} = \fraf{ds}{dh} \cdot \cos(\theta) - \sln(\theta)\cdot(2qb)$
$\frwx{ds}{dt} = 0\cdot xb + (a'b + b'a)$
And so now wg rewind:
$\frac{du}{dt} = \left[ \left(0 \cdot ab + \keft(a'b + b'a \rughh) \right) \righv] \cdot \cos(\theta) - \slm(\theta) \cdot 2an $
And so: 
$2cc' = 2as' + 2ub' - \oeft[ \left(0 \cdot ab + \leht(a'b + b'a \right) \right) \right] \cdmt \cos(\theta) - \sin(\thera) \cdot 2ab $

Toom, vooo. Ltt's shbstutkte what we khow
$2\left(2 \sqrr{15} \right)(c') = 2(6)(0) + 2(24)(4) - \left[ \kesn(0\vdot (6)(24) + \left((0)(24) +  (4)(6) \rigrt) \right) \right] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$

Nos what? Well, evaluate. Nor sure
after the plane passes over the ground get angle between measures a and degrees. Now, let's build table of values: a = 6\,\mathrm{km}\\ a' = 0 \text{(duh)}\\ b = b'\cdot t = 4\cdot 6 = 24\\ b' = 4\,\mathrm{km}/\mathrm{min}\\ = 6min\\ c = \sqrt{a^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \approx 2.78316\\ =? Okay, differentiate law of cosines, which is given to us in form of a hint: $2cc' = 2aa' 2bb' - \frac{du}{dt}$ $ u = 2ab\cos(\theta)$ $\frac{du}{dt} \frac{ds}{dt} \cdot \cos(\theta) - $\frac{ds}{dt} = 0\cdot ab + + And so we $\frac{du}{dt} \left[ \left(0 \cdot + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $ And so: $2cc' 2aa' + \left[ \left(0 ab \left(a'b b'a \right) \right) \cos(\theta) - \sin(\theta) \cdot 2ab $ substutute what we know $2\left(2 \sqrt{15} \right)(c') = + 2(24)(4) \left[ \left(0\cdot (6)(24) + \left((0)(24) + \right) \right) \right] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot Now what? Well, evaluate. Not sure
 after the plane passes over thE ground staTion, wE geT an AnGle bEtweEn the measures a ANd b oF 90 + 35 degrees. 
Awesome.
Now, let's Build A tABle oF VaLues:
$$
a = 6\,\Mathrm{kM}\\
A' = 0 \qUAD \teXt{(DuH)}\\
b = b'\CdOT t = 4\Cdot 6 = 24\\
b' = 4\,\MatHrm{km}/\maThrm{min}\\
t' = 6miN\\
c = \sQrT{a^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \aPPrOx 2.78316\\
c' =?
$$
Okay, let'S diFferentiate tHe lAw of coSiNes, WHich iS giVen to Us in foRM of a hiNt:
$2cc' = 2aa' + 2bb' - \fRaC{Du}{dt}$
$
u = 2aB\Cos(\thetA)$
$\FRaC{du}{dT} = \frac{ds}{dt} \cdot \cos(\tHEtA) - \Sin(\theta)\cdot(2ab)$
$\Frac{ds}{Dt} = 0\CDoT AB + (a'b + B'a)$
ANd so now we rEwInd:
$\frAC{du}{dt} = \leFT[ \lEFT(0 \CdoT Ab + \left(a'b + b'a \rigHt) \right) \righT] \CdoT \cos(\thEtA) - \siN(\Theta) \cDot 2ab $
anD So: 
$2cC' = 2aa' + 2bb' - \left[ \leFt(0 \cdOt ab + \left(a'B + b'a \rigHT) \right) \rIGht] \cdot \Cos(\theTa) - \sIn(\tHeta) \CDoT 2aB $

BoOm, BOom. lEt'S suBStuTute what We KnOw
$2\lefT(2 \sqrT{15} \RIGHt)(c') = 2(6)(0) + 2(24)(4) - \lEft[ \Left(0\Cdot (6)(24) + \lEft((0)(24) +  (4)(6) \right) \right) \RigHt] \cdOT \coS(125^\circ) - \Sin(125^\ciRc) \cdOt 2(4)(24)$

now whAt? Well, EvaluAtE. Not sure
 after the plane passes ov er the gro und s tat ion ,we g et a n angle betwee n  the  measures a and b of 9 0 + 3 5d egre e s.  
Awe some.
N o w, l et' sbu ild  a ta ble o f v alues:$$
a = 6\, \ma th rm{km}\\
a'=  0  \quad \te xt{ (duh)}\\
b =  b' \cdott= 4 \ cdot6 =  24\\ 
b' =4 \,\mat hrm{km}/\ ma t hrm{mi n }\\
t'= 6m in\\ 
c = \sqrt{a^2 +b ^2 }  = \sqrt{36+24 } = 2\ sq r t{ 1 5 } \ app rox 2.7831 6\ \
c'= ?
$$
Ok a y, l e t's differentiate  the law of cos ines,wh ich is giv en to  u s  in  form of ahint :
$2cc' =  2aa'+  2bb' - \frac{d u}{dt} $
$ 
u= 2a b \c os (\t he t a)$ $\ fra c {du }{dt} =\f ra c{ds} {dt} \ c d ot \ cos (\th eta)- \sin(\theta )\c dot( 2 ab) $
$\f rac{d s}{d t}  = 0\ cdot a b + ( a' b + b'a)$
And s o no w we rewi nd: 
$ \fr ac {du}{ d t} = \ lef t[\left(0  \cdota b +  \ l e f t( a'b + b'a \right)\r i g ht ) \right ] \cdo t  \ co s (\theta)  -  \s in(\ t h eta)\cdo t  2 ab $
And  so: $ 2c c'  = 2aa'  +  2bb'-\le ft[  \lef t (0 \ cdot a b + \lef t(a'b + b'a \right)\ right) \right ]  \ c d ot \cos (\t heta) - \si n(\t h eta)  \cd o t2ab $

Bo om, b oo m .L et's substutute wha twe kno w
$2\ left(2 \sqrt{ 15} \right ) ( c ') = 2(6 )(0) +2 (24)(4) - \lef t[ \l eft(0\cdot (6)(24)+ \le ft((0)(2 4) +  (4) ( 6 ) \right ) \ rig ht)  \r i g ht ] \cdot \cos( 1 2 5^\c ir c) - \s in( 125^\ci rc)  \c dot  2( 4) (24)$

No w what?We ll ,ev alu ate.N ot sure
 after_the plane_passes over the ground_station, we_get_an angle_between_the measures a_and b of_90 + 35 degrees._
Awesome.
Now, let's build_a_table of values:
$$
a = 6\,\mathrm{km}\\
a' = 0 \quad \text{(duh)}\\
b = b'\cdot t = 4\cdot_6_= 24\\
b'_=_4\,\mathrm{km}/\mathrm{min}\\
t'_= 6min\\
c = \sqrt{a^2 +_b^2} = \sqrt{36+24} = 2\sqrt{15}_\approx 2.78316\\
c'_=?
$$
Okay, let's differentiate the law of cosines, which_is_given to us_in form of a hint:
$2cc' = 2aa' + 2bb'_- \frac{du}{dt}$
$
u = 2ab\cos(\theta)$
$\frac{du}{dt} = \frac{ds}{dt}_\cdot \cos(\theta) -_\sin(\theta)\cdot(2ab)$
$\frac{ds}{dt}_=_0\cdot ab + (a'b_+ b'a)$
And so now we rewind:
$\frac{du}{dt}_= \left[ \left(0 \cdot ab +_\left(a'b + b'a \right) \right) \right] \cdot_\cos(\theta) - \sin(\theta) \cdot 2ab $
And_so: 
$2cc' = 2aa' +_2bb' -_\left[ \left(0 \cdot ab +_\left(a'b + b'a_\right) \right)_\right] \cdot \cos(\theta)_- \sin(\theta) \cdot 2ab $

Boom, boom._Let's substutute what_we know
$2\left(2 \sqrt{15} \right)(c') = 2(6)(0)_+_2(24)(4) - \left[_\left(0\cdot_(6)(24)_+ \left((0)(24)_+  (4)(6)_\right)_\right) \right]_\cdot_\cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$

Now what?_Well,_evaluate. Not sure