I usually just use sine rule but it doesnt work when theres unknown values
And so far I got arc length = 2a and sector area = 2a cm^2
#can someone explain how to do this
red arch
last oreBOT
I usually just use sine rule but it doesnt work when theres unknown values
And s...
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tender ledge
I usually just use sine rule but it doesnt work when theres unknown values
And s...
Let's try it generally.
Suppose AD = r, AB = R. Then, as you said:
l(DE) = αr
S(ADE) = (1/2)αr^2
Next, try finding the area of ABC and the lengths of CD and BE.
red arch
I got stuck on the area of ABC
how can I get it if I dont have CB
i either use the 1/2absinC or 1/2basexheight
tender ledge
Well, let's see.
S(ABC) = (1/2)AB*BC
We know that AB = R. How can you find BC, knowing R and α?
tender ledge
Yes. In general:
tan(α) = BC/AB
So, BC = AB tan(α) = R tan(α).
Hm, let me check.
S(ABC) = (1/2)AB*BC = (1/2)R*R tan(α) = (1/2)R^2 tan(α)
S(shaded) = S(ABC) - S(ADE) = (1/2)(R^2 tan(α) - αr^2) = (1/2)(16tan(α) - 4α) cm^2 = (8tan(α) - 2α) cm^2
So yeah, correct.
What about the perimeter?
tender ledge
Yup.
tender ledge
Hm, let's see. We have:
P(shaded) = ED + CD + BC + BE = αr + (R/cos(α) - r) + R tan(α) + (R - r) = (1/cos(α) + tan(α) + 1)R + (α - 2)r
Not sure if you simplified the terms with r correctly. Maybe you forgot to add something?
tender ledge
Ah, wait. I didn't simplify the trigonometric expression correctly, one sec.
tender ledge
Hm, let's see. We have:
P(shaded) = ED + CD + BC + BE = αr + (R/cos(α) - r) + R ...
Ok, it's good now.
tender ledge
so 2a + 4/cos(a) + 4tan(a)
Yeah, that seems good.
red arch
that was the first quesiton in about 3 years where I used tan and cos...
I relied solely upon sine rule 😭
thanks for explaining :)
tender ledge
You're welcome!
red arch
+close