BCDE is a square. TSD and CSE are straight lines. CB = CT and ∠SCT = 13°. Find
- ∠CBT
- ∠EDT
.
(a)
∠BCE = 45° (Right angle)
∠BCT
= ∠BCE - ∠SCT
= 45° - 13°
= 32°
∠CBT
= (180° - ∠BCT) ÷ 2
= (180° - 32°) ÷ 2
= 148° ÷ 2
= 74° (Isosceles triangle)
(b)
CB = CT = CD
TCD is an isosceles triangle.
∠CTD = ∠CDT (Isosceles triangle)
∠CDT
= (180° - ∠ECD - ∠SCT) ÷ 2
= (180° - 45° -13°) ÷ 2
= 122° ÷ 2
= 61°
∠EDT
= ∠CDE - ∠CDT
= 90° - 61°
= 29°
Answer(s): (a) 74°; (b) 29°
