Qianqi Zhao's answer to Chong Jia Tao's Secondary 3 E Maths Singapore question.
done
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Idk if I’m correct lol but here’s my ans
Date Posted:
4 years ago
First one not correct. Question asks for its sine and the working should NOT include the calculated value of the angle (or even write it out, so in nowhere should ACD be seen at all).
Correct is
sin ACB = opp / hyp = 6 / 10
sin ACD
= sin (180 - ACB)
[because ACD + ACB = 180]
= sin ACB
[by the supplementary angle rule sin (180 - x) = sin x]
= 6/10
= 3/5
sin ACB = opp / hyp = 6 / 10
sin ACD
= sin (180 - ACB)
[because ACD + ACB = 180]
= sin ACB
[by the supplementary angle rule sin (180 - x) = sin x]
= 6/10
= 3/5
Second one is a hence question, so method is not accepted as well.
Area of triangle ACD
= 1/2 * AC * CD * sin ACD
= 1/2 * 10 * 3 * 0.6
(the “hence” is used here since sin ACD is directly substituted with 0.6)
= 9 cm^2
Area of triangle ACD
= 1/2 * AC * CD * sin ACD
= 1/2 * 10 * 3 * 0.6
(the “hence” is used here since sin ACD is directly substituted with 0.6)
= 9 cm^2
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