Illuminati?

$Inradius~ (r)~of~ a~ triangle~=~\dfrac{\Delta}{s}$ $where~\Delta~=~area~of ~the~triangle~and~(s)~is ~semiperimeter$

$\Rightarrow r=\dfrac { \sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } }{ s }$

$\Rightarrow r=\sqrt { \dfrac { \left( s-a \right) \left( s-b \right) \left( s-c \right) }{ s } }$

$where~a=6+8=14~,~b=x+8 ~and~c=6+x$

$And~ Given ~that ~inradius ~of~ the ~given~ triangle (r)~~=4~cm$

$\Rightarrow 4=\sqrt { \dfrac { \left( 14+x-14 \right) \left( 14+x-\left( 8+x \right) \right) \left( 14+x\left( 6+x \right) \right) }{ 14+x } }$

$\Rightarrow 16=\dfrac { \left( x \right) \left( 6 \right) \left( 8 \right) }{ 14-x }$

$\Rightarrow 224-16x=48x$

$\therefore x=\dfrac { 224 }{ 32 } =7$

tan θ = 4/6 = 2/3

tan 2θ = 2(2/3)/(1-(2/3)^2)=12/5

implies sin 2θ = 12/13 and cos 2θ = 5/13

tan α = 4/8 = 1/2

tan 2α = 2(1/2)/(1-(1/2)^2)=4/3

implies sin 2α = 4/5 and cos 2α = 3/5

sin ∠BAC = sin (180-(2α+2θ))

=sin(2α+2θ)

=sin(2α)cos(2θ)+cos(2α)sin(2θ)

=(4/5)(5/13)+(3/5)(12/13)

=56/65

Using sine theorem:

(x+8)/sin 2θ=(6+8)/sin ∠BAC

(x+8)/(12/13)=14/(56/65)

=> x=7

Proof of my assumptions :

In triangle BDO and BEO,

DO=OE=radius=4

∠BDO=∠BEO=90° (Tangents to the circle)

BO is common in both triangles.

Therefore, triangle BDO and BEO are congruent. (RHS)

So, we can say ∠DBO=OBE=θ.

Similarly, triangle CEO and COF are congruent.

Consider the above representation of the problem. The red lines connect the center of the circle to the vertices of the triangle. Notice that by circle theorems, they bisect the coloured regions, split by the radius of the triangle, into two congruent parts.

By trigonometric functions: $\tan{\alpha} = 4/6 \\ \tan{\beta} = 4/8 \\ \tan{\gamma} = 4/x$ As angles in a triangle add to 180 degrees: $2\alpha + 2\beta + 2\gamma = 180 \\ \arctan{4/6} + \arctan{4/8} + \arctan{4/x} = 90$ $\tan(\arctan{4/6} + \arctan{4/8} + \arctan{4/x})= udf \tag{*}$

I'll now use the identity $\tan(a+b+c)= \frac{\tan a + \tan b + \tan c - \tan a \tan b \tan c}{1 - \tan a \tan b - \tan b \tan c - \tan c \tan a}$ We can plug in the numbers into this identity but if we first notice that since the right hand side of (*) was undefined, the denominator of the identity must be zero so therefore $1 - \tan (\arctan 4/6) \tan (\arctan 4/8) - \tan (\arctan 4/8) \tan (\arctan 4/x) - \tan (\arctan 4/x) \tan (\arctan 4/6) = 0 \\ 1 = 4/6 * 4/8 + 4/8 * 4/x + 4/x * 4/6 \\ 1 = 1/3 + 2/x + 8/3x \\ 2/3 = 6/3x + 8/3x \\ 2 = 14/x \\\boxed{x = 7}$

tan(B/2)=4/6=2/3

tanB=2tan(B/2)/(1-tan²B/2)=12/5

tan(C/2)=4/8=1/2

tanC=4/3

A+B+C=180°

tan(B+C)=tan(180°-A)=-56/33=-tanA

tan(A/2)=4/x

tanA=(8/x)(1-16/x²)=56/33

7x²-33x-112=0

x=7

The semi perimeter of the triangle is (14+x). We know ,inradius of circle inscribed in a triangle = area of triangle / semi perimeter i.e. r = area /s ........(1). Here , r= 4 , area by heroin's formula is sq rt s(s-a).(s-b).(s-c) Substituting the values in (1) and squaring nd solving we get x= 7

Since the sides of △ABC are tangents of the incircle, the lengths of AB, BC, CA are x+6, 14, and x+8 respectively.

So when we apply the cosine rule, we can get an equation:

(x+8)^2 = (x+6)^2 + 14^2– 2(x+6) 14 cos ∠ABC

Now note that ∠ABC = ∠ABO + ∠OBC when O is the center of incircle.

And ∠ABO = ∠OBC because tan ∠ABO = tan ∠OBC = r/6 = 4/6

So cos ∠ABC = cos 2(∠OBC )

= cos2 ∠OBC - sin2 ∠OBC = (36-16)/52 = 5/13

No plug in for cos ∠ABC = 5/13, we’ll solve for x:

(x+8)^2 = (x+6)^2 + 14^2– 2(x+6) 14 (5/13)

X2+ 16x +64 = X2+ 12x +36 + 196 – (140/13)(x+6)

(140/13)(x) + (140*6/13) = 168 – 4x

(192/13)(x) = 168 - (140*6/13)

192x = 13 (168) - (140) 6 = 1,344

X = 7

use the fact that area of the triangle can be found out by heron's formula as well adding the area of 3 triangles which comprises the whole triangle each having the height 4cm and bases as 14cm, (x+8)cm, (x+6)cm. thus equating the area of triangle ABC found in 2 methods we have : {(14+x) (x) (6) (8)}^(1/2) = ½{(4) (14) + (4 (8+x)) + (4 (6+x))} solving this equation we get the equation {(x^2) +7x-98} = 0 solving it we get x = 7.

Project a perpendicular line from center of the circle to member BC it will intersect with the tangent point call it Z ,,call the tangent point with member AC `N' & center of the circle 'M' ∆ czm is right angle so mc=√16+64=4√5, Angle {mcz}={mcn}=P so cos{P}=8/4√5=2÷√5 , Angle ACB=2p=U so cos{U}=2×{4÷5}-1=(3÷5). By using the law of cosin, cos(U)=3÷5=14^2+(8+X)^2-(6+X)^2/(2×14×(8+X) So 3×28×(8+X)=5×(196+64-36+4X) so 64X=448,,, X=7########

OE⊥AC,and assume the extended center lind AO⊥BC at D，then △ADC≌△AEO.we use te Pythagoras theorem can estimates the X about 7.5,so the answer is 7 or 8,so,if the X=8,must have BE⊥BC,but they don`t,so the answer must is 7

I had solve it by using the law of sines

Isn't this class 10 ncert maths?

I kind of brute forced it by using x in equations. I found the area of the triangle with Heron's Formula and used the product of the inradius and semiperimeter to get the area again. This gave me: sqrt((x+14)(x)(6)(8))=56+4x 48(x+14)(x) = 16(x+14)^2 3x = x+14 2x = 14 x = 7 I then solved this to get one real solution, x = 7.

Consider the well-known 13-14-15 triangle. Note that its inradius is well-known to be 4. Since $BC = 14$ , it follows that $15=8+x$ (or $13 = 6+ x$ ) thus $\boxed{x=7}$ .

Area of the triangle made of the vertices B,C and incentre= 1/2x4x(6+8)=28. therefore area of the triangle ABC (delta)= 3*28= 84. As we know that, r=delta/s. now s=((6+8)+(8+x)+(x+6))/2=14+x. Delta=84, r=4 Therefore, s=14+x=84/4=21. so, x=7.