In the diagram, ADC is a straight line. AB = 11 cm, BD = 10 cm and CD = 3 cm. Using as 
much of the information given below as is necessary, calculate
(i) sin ∡BAD,
(ii) BC²,
(iii) the area of triangle BCD
given that sin ∡ADB = 0.88, cos ∡ADB = -0.47, tan ∡ADB = -1.88
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Answer:
(i) Using sine rule,
sin ∡BAD / 10 = sin ∡ADB / 11
sin ∡BAD = 0.88 / 11 * 10
sin ∡BAD = 0.8
(ii) cos ∡BDC = - cos ∡ADB = 0.47
Using cosine rule,
BC² = BD² + CD² - 2 (BD)(CD)(cos ∡BDC)
BC² = 10² + 3² - 2(10)(3)(0.47)
BC² = 80.8
(iii) sin ∡BDC = sin ∡ADB = 0.88
Area of triangle BCD
= ½(BD)(CD) sin ∡BDC
= ½(10)(3) (0.88)
= 13.2 cm²
