Yxcas's answer to >_< 's Secondary 3 E Maths Singapore question.
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Let line CE be a line which CE // DA and CE = DA making AECD a parallelogram.
Using properties of a parallelogram,
∠DAB = ∠CEB and DC = AE
Hence,
∠BCE = 180 - 76 - 60 = 44
EB = 18 - 10 = 8
Using Sine rule,
BC / sin∠CEB = EB / sin∠BCE
BC / sin76° = 8 / sin44°
BC = 8 / sin44° * sin76°
BC = 8 / sin44° * sin76°
BC = 11.17cm (2d.p.)
Using properties of a parallelogram,
∠DAB = ∠CEB and DC = AE
Hence,
∠BCE = 180 - 76 - 60 = 44
EB = 18 - 10 = 8
Using Sine rule,
BC / sin∠CEB = EB / sin∠BCE
BC / sin76° = 8 / sin44°
BC = 8 / sin44° * sin76°
BC = 8 / sin44° * sin76°
BC = 11.17cm (2d.p.)
Date Posted:
2 months ago
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