Question

solve for x.


#happy earth day​

Answer 1

$\huge{\orange{\fbox{\green{\fbox{\mathcal{\orange{An}\blue{sw}\green{er}}}}}}}$

$\large{\orange{\fbox{\green{\fbox{\mathcal{\orange{No}\green{te}}}}}}}$

$\large{\orange{\fbox{\green{\fbox{\mathcal{\orange{Me}\blue{th}\green{od}:-1}}}}}}$

$\large{\red{\fbox{\green{Calculate\; some\; known \;angles}}}}$

• ACB = 180-(10+70)-(60+20) = 20°
• AEB = 180-70-(60+20) = 30°

Draw a line from point D parallel to AB, labeling the intersection with BC as a new point F and conclude:

∆DCF and ∆ACB

• CFD = CBA = 60+20 = 80°
• DFB = 180-80 = 100°
• CDF = CAB = 70+10 = 80°
• ADF = 180-80 = 100°
• BDF = 180-100-20 = 60°

Draw a line FA labeling the intersection with DB as a new point G and conclude:

∆ADF and ∆BFD

• AFD = BDF = 60°
• DGF = 180-60-60 = 60° = AGB
• GAB = 180-60-60 = 60°
• DFG (with all angles 60°) is equilateral
• AGB (with all angles 60°) is equilateral

CFA with two 20° angles is isosceles, so FC = FA

Draw a line CG, which bisects ACB and conclude:

• ACG CAE
• FC-CE = FA-AG = FE = FG
• FG = FD, so FE = FD
• With two equal sides, DFE is isosceles and conclude:
• DEF = 30+x = (180-80)/2 = 50

• x = 20°

$\large{\orange{\fbox{\green{\fbox{\mathcal{\orange{Me}\blue{th}\green{od}:-2}}}}}}$

• Construct a triangle practically with the required measurements and the find out the value of x