Right Triangle Lengths

Let $\angle CDP = \theta$ and $CP = x$ .

Then $\tan(\theta) = \dfrac{x}{\sqrt{3}}$ and $\tan(\theta + 30) = \dfrac{x + 2}{\sqrt{3}}$ .

Now $\tan(\theta + 30) = \dfrac{\tan(\theta) + \tan(30)}{1 - \tan(\theta)*\tan(30)}$ , so

$\dfrac{x + 2}{\sqrt{3}} = \dfrac{\frac{x}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - (\frac{x}{\sqrt{3}})*(\frac{1}{\sqrt{3}})} \Longrightarrow x + 2 = \dfrac{x + 1}{1 - \frac{x}{3}}$

$\Longrightarrow (x + 2)(3 - x) = 3*(x + 1) \Longrightarrow x^{2} + 2x - 3 = 0 \Longrightarrow (x + 3)(x - 1) = 0$ .

Now since we require that $x \gt 0$ we conclude that $x = \boxed{1}$ .

Let B = CDP and X = CP:

Tan B = X/sqrt(3) Tan (30 + B) = (2 + X)/sqrt(3) By trigo identities; Tan (M +- N) = (Tan M +- Tan N)/(1 -+ Tan M Tan N) Tan 30 = 1/sqrt(3) = sqrt(3)/3 So, (Tan 30 + Tan B)/(1 - Tan 30 Tan B) = (2 + X)/sqrt(3) Subs Tan B = X/sqrt 3 Solving the equation will give X = 1

Consider triangle PDQ. By the law of cosines:

$DP^2+DQ^2-2 \cos 30° DP \cdot PQ = 4$

By Pithagoras we have:

$DP^2= CP^2 + 3$

and

$DQ^2 = (CP + 2)^2+3$

As for the product of DQ and DP, let us consider the area A of triangle DPQ:

$\dfrac{A}{2} = 2 \sqrt{3} = DQ \cdot DP \cdot \sin 30°$

hence

$DP \cdot DQ = 4 \sqrt{3}$

Let us substitute in the first equation and solve by CP:

$CP^2+3+CP^2+2CP+4+3 - 2 \cdot 4\sqrt{3} \cdot \dfrac{\sqrt{3}}{2} = 4$

$\iff 2CP^2+2CP -6 = 0$

$\iff CP^2+CP-3 = (CP+3) (CP-1) =0$

and since CP cannot be -3, it must be 1.

Let B = CDP and X = CP:

Tan B = X/sqrt(3) Tan (30 + B) = (2 + X)/sqrt(3) By trigo identities; Tan (M +- N) = (Tan M +- Tan N)/(1 -+ Tan M Tan N) Tan 30 = 1/sqrt(3) = sqrt(3)/3 So, (Tan 30 + Tan B)/(1 - Tan 30 Tan B) = (2 + X)/sqrt(3) Subs Tan B = X/sqrt 3 Solving the equation will give X = 1

area of∆ dcq=1\2×(2+R)×√3=area∆ cpd+area ∆pqd=1\2×R×√3+1\2×M×N×sin30,R=cp,M=Dp,N=DQ,OK ,,1\2×(2+R)√3=1\2×R×√3+1\2×M×N×1\2,,,so 2√3+R√3=R√3+M×N×1\2,M×N=4√3,,,,,,,,,,,,,,,, cos30=M^2+N^2-4\2M.N ,√3×M.N=M^2+N^2-4. M×N=4√3 ,so 4√3×√3=M^2+48\M^2-4. M^2+48\M^2=16,multiple expression by M^2. M^4-16M^2+48=0.(m^2-12).(m^2-4)=0 so m^2=12 Or 4 , m^2. N^2=48 so ,N^2=48\12=4 . N=2 rejected N>PQ>2.,,,so M^2=2 , . R^2=2^2-√3^2=1.R=1=CP###########

Length of PQ=length of PD By using Pythagora's thm PD^{2}=DC^{2}+CP^{2} thus to get the length of CP, CP=\sqrt{2^2-3} so CP=1

triangle DCP is a special triangle... a 30-60-90 triangle whose sides are always x-x*3^1/2-2x... once you see that the question pretty much solves itself.

put, CP=X, area of CDQ(1/2 x root 3 x X+2) = area CDP(1/2 x root 3 x X)+ area DPQ ( 1/2 x sin 30 x DP x DQ)...........(1)

here, DP = root (3+x^2) and DQ = root(3+ (X+2)^2) solve eqn (1), the possible value of it comes 1, hence CP=1

triangle DPQ having base-2 m and perpendicular height-root3 .then
tan 30=(2+x)/root3; 1/root3=(2+x)/root3; 2+x=1. hence x=-1(where x is length of CP) solength of CP=1

Triangle DCQ is a 30-60-90 triangle since angle CDQ = 60 and angle C = 90. Since angle Q = 30, triangle PDQ is isosceles which tells us that DP = 2m. Now use the Pythagorean theorem on triangle CDP to find CP: CD^2+CP^2=DP^2 3 + CP^2 = 4

which tells us CP = 1

Triangle DCQ is a 30-60-90 triangle since angle CDQ = 60 and angle C = 90. Since angle Q = 30, triangle PDQ is isosceles which tells us that DP = 2m. Now use the Pythagorean theorem on triangle CDP to find CP:

3 + CP^2 = 4

which tells us CP = 1

Of course there's a rigorous method to find the length but here's what i did- i randomly chose the value of angle CPD, first 30 but that means CDQ is 90 that's absurd. Then i put CPD=60, that means CQD=30=>DP=PQ=2 & CP=1 & i checked this combination using Pythagoras and it worked!! Pretty much saved my time.

I made system with , Pitagora' s theorem , Stweart's theorem and law of cosine, solve it to find DQ , CP and DP