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Gemiola [76]
2 years ago
5

In right ∆ABC with m∠B=30°, AC = 4. Find AB, HB and Area of triangle ABC

Mathematics
2 answers:
alexdok [17]2 years ago
6 0

Answer:

hb=4√3

Step-by-step explanation:

iogann1982 [59]2 years ago
5 0

Answer:

AB = 8, HB = 6, Area of ∆ABC = 8\sqrt{3}, Perimeter of ∆ABC = 12 + 4\sqrt{3}

Step-by-step explanation:

To find AB:

∆ABC is an 30,60,90∆

Using the theorem, you can find AB = 2AC = 2*4 = 8

AB = 8

To find HB:

You need to find AH to subtract from AB

Construct CH, a perpendicular bisector to side AB

From before you can put together that m∠CAB = 60°

∆ACH is an 30,60,90∆

Using this method again, AH = AC/2 = 4/2 = 2

Then you subtract AH from AB = 8-2 = 6

HB = 6

To find the area of ∆ABC:

You use the (base*height)/2 method

base = AB = 8

to find the height, CH, you need to use the Pytha Theorem

and get AH^{2}+CH^{2}=AC^{2}

then substitute, and get 2^{2} + CH^{2} = 4^{2}

calculate and get CH = 2\sqrt{3}

then the height = CH =  2\sqrt{3}

solve the area and get

Area of ∆ABC = 8\sqrt{3}

(optional perimeter)

to find perimeter of ∆ABC:

you add AC + CB + AB

you find CB by using opposite to 30°

CB = CH*2 = 2\sqrt{3}*2 = 4\sqrt3}

so AC + CB + AB = 4 + 4\sqrt3} + 8

Perimeter of ∆ABC = 12 + 4\sqrt3}

Hope this helps!!

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podryga [215]

Answer:

B

Step-by-step explanation:

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Suppose that the proportions of blood phenotypes in a particular population are as follows: A B AB O 0.48 0.13 0.03 0.36 Assumin
Serggg [28]

Answer:

P(O and O) =0.1296

P=0.3778

Step-by-step explanation:

Given that

blood phenotypes in a particular population

A=0.48

B=0.13

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As we know that when A and B both are independent that

P(A and B)= P(A) X P(B)

The probability that both phenotypes O are in independent:

P(O and O)= P(O) X P(O)

P(O and O)= 0.36 X 0.36 =0.1296

P(O and O) =0.1296

The probability that the phenotypes of two randomly selected individuals match:

Here  four case are possible

So

P=P(A and A)+P(B and B)+P(AB and AB)+P(O and O)

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7 0
3 years ago
Is this statement true or false? a^(-3)+a^(-3)=a^(-6)​
EleoNora [17]

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I'm being timed pls help will pick brainliest
stiv31 [10]

Answer:

Cos 0° = 6.42 = V1

6th quadrant.

So, cos x = 0 implies x = (2n + 1)π/2 , where n takes the value of any integer. For a triangle, ABC having the sides a, b, and c opposite the angles A, B, and C, the cosine law is defined. In the same way, we can derive other values of cos degrees like 30°, 45°, 60°, 90°, 180°, 270°and 360

Step-by-step explanation:

if sin a= 3/4 then a = 50

if sin = 4/5 then a = 60

if sin = 2/3 then a = 40

But we can perfect this

if sin = 4/5 then a = 57   as 4/5 = 0.83

We want 0.8 = 4/5

if sin = 4/5 then a = 54 as 4/5 = 0.809

if sin = 4/5 then a = 53.5 as 4/5 = 0.80385

Now for cos

It is much easier than it initially appears. Remember the definition of SINE:

SINΘ = opp             In your case, that means the opposite is 4/5 = 0.80385 (yes, ignore the sign for now) and the hypotenuse is 11.48910018

           hyp             Please draw that triangle right now, because it will help you a lot at the end.

 53.50 degree           Remember to place the angle in the appropriate spot.

Now, use the Pythagorean Theorem to find the missing side (easy, right? It's 9.534) and place it in the adjacent position.

You can easily find all of the trig functions now!

Simply remember that:

COSΘ = Adj                                  with SEC the reciprocal of this one

            Hyp

TANΘ = Opp                                  COT the reciprocal of TAN and, if anyone asks, CSC coming from SIN.

            Adj

I told you to ignore the signs, but now we can't anymore. Remember the four quadrants and the memory trick:

A                        -- ALL are positive

Smart                 -- SIN and CSC are positive

Trig                    -- TAN & COT are positive

Class                  -- COS and SEC are positive.

Since your SIN was negative, it must be in III or 6, and COS is positive in I and 6 So we're in quadrant 6 then!

Only your COS and SEC will be positive, the rest negative

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sleet_krkn [62]

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