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I'm still trying out question 9, but here's the answer to question 10 first. Hope it helps.
Date Posted:
3 years ago
For question 9,
Similar to question 10, use the formula
(sin A / a) = (sin B / b)
sin ABD / AD = sin ADB / AB
sin 25° / 4 = sin ADB / 8
sin ADB = 8 × sin 25° / 4
ADB = sin^-1(8 × sin 25° / 4)
Now you can find the angle of ACD, ACB and BAC.
You can use the same method to find question b and c. Or you can use cosine rule:
a² = b² + c² - 2bc cosA
Similar to question 10, use the formula
(sin A / a) = (sin B / b)
sin ABD / AD = sin ADB / AB
sin 25° / 4 = sin ADB / 8
sin ADB = 8 × sin 25° / 4
ADB = sin^-1(8 × sin 25° / 4)
Now you can find the angle of ACD, ACB and BAC.
You can use the same method to find question b and c. Or you can use cosine rule:
a² = b² + c² - 2bc cosA
urmm but the answer i got from calculator is 57.6972... while the answer on my qn paper is 57°
Hi Sally,
Maybe check with your teacher if the answer is wrong? Or maybe the answer is rounded off. Because if let's say 57° is correct, we can use sine rule to show:
sinADB / AB = sinABD / AD
sin57° / 8 = sin25° / 4
0.104833821 = 0.1056545654
Since both sides are not equal, 57° is inaccurate.
Maybe check with your teacher if the answer is wrong? Or maybe the answer is rounded off. Because if let's say 57° is correct, we can use sine rule to show:
sinADB / AB = sinABD / AD
sin57° / 8 = sin25° / 4
0.104833821 = 0.1056545654
Since both sides are not equal, 57° is inaccurate.