ABCD is a square. DCC and BCD are straight lines. BA = BD and ∠CBD = 13°. Find
- ∠BAD
- ∠DCD
.
(a)
∠ABD = 45° (Right angle)
∠ABD
= ∠ABD - ∠CBD
= 45° - 13°
= 32°
∠BAD
= (180° - ∠ABD) ÷ 2
= (180° - 32°) ÷ 2
= 148° ÷ 2
= 74° (Isosceles triangle)
(b)
BA = BD = BC
DBC is an isosceles triangle.
∠BDC = ∠BCD (Isosceles triangle)
∠BCD
= (180° - ∠DBC - ∠CBD) ÷ 2
= (180° - 45° -13°) ÷ 2
= 122° ÷ 2
= 61°
∠DCD
= ∠BCD - ∠BCD
= 90° - 61°
= 29°
Answer(s): (a) 74°; (b) 29°
