The below figure represent the schematic diagram of the system.

       Figure -(1)

Write the expression of cosine rule in ΔADC for the angle ∠ADC.

 cos∠ADC=AC2−AD2−DC22(AD)(DC) ......... (I)

Here, the sides of ΔADC is AD, DC and AC respectively.

Write the expression for the relation between angles in ΔADC.

 ∠ACD=∠CAD ...... (II)

Write the expression of sum of angles in ΔADC .

 ∠ACD+∠CAD+∠ADC=180°∠CAD+∠CAD+∠ADC=180°2∠CAD+∠ADC=180° ...... (III)

Write the expression of ∠DAB.

 ∠DAB=∠BAC−∠CAD ...... (IV)

Write the expression for the relation between angles in ΔABD.

 ∠DAB=∠DBA ...... (V)

Write the expression of ∠ADB.

 ∠ADB=180°−∠DAB−∠DBA=180°−∠DAB−∠DAB=180°−2∠DAB ...... (VI)

Write the expression of cosine rule in ΔADB.

 AB2=AD2+AB2−2(AD)(AB)cos∠ADB ......... (VII)

Calculation:

Substitute 40° for ∠ADC, 0.750 in for AD, 0.750 in for DC in Equation (I).

 cos40°=AC2− ( 0.750 in )2− ( 0.750 in )22( 0.750 in)( 0.750 in)AC2−(0.750 in)2−(0.750 in)2=(0.8618  in2)AC2=0.2632 in2AC=0.513 in

Substitute 40° for ∠ADC in Equation (III).

 2∠CAD+40°=180°2∠CAD=180°−40°∠CAD=140°2∠CAD=70°

Substitute 116° for ∠BAC and 70° for ∠CAD in Equation (IV).

 ∠DAB=116°−70°=46°

Substitute 46° for ∠DAB in Equation (VI).

 ∠ADB=180°−2(46°)=180°−92°=88°

Substitute x for AB, 88° for ∠DAB, 0.750 in for AD, 0.750 in for AB in Equation (VII).

 x2=(0.750 in)2+(0.750 in)2−2(0.750 in)(0.750 in)cos88°x2=1.0857 in2x=1.0857  in2x=1.041 in

Conclusion:

The distance x is 1.041 in.