Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=> [tex]\frac{180-x}{2}+45=115 \\[/tex]
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) => [tex]\left \{ {{y=x} \atop {2x-y=40}} \right.\\\\=> \left \{ {{x=40} \atop {y=40}} \right.[/tex]
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°
