Olympiad Math | Calculate the length AD in the Triangle ABC | Important Geometry skills explained

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Olympiad Math | Calculate the length AD in the Triangle ABC | Important Geometry skills explained

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Calculate the length AD

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[bigm]

I can’t sleep until I’ve watched the latest Premath!🥂❤

[jimlocke]

At about 2:50, drop a perpendicular from D to AB and label the intersection as point E. Note that ΔADE is the special 30°-60°-90° right triangle with hypotenuse of length x, so length DE = x/2 and AE = (x√3)/2. Also note that ΔABC and ΔEBD are similar right triangles with long side twice as long as short side, so, because length DE = x/2, length BE = x. Then, length AE = 30-x. Equating the 2 values for length AE, 30-x=(x√3)/2. Solving, x=30/(1+(√3)/2) or x=60/(2+√3), as PreMath also found.

[Copernicusfreud]

Yay! I solved it. I calculated all the interior angles of the lines of the triangles and then used the law of sines to find the length of AD. I started out with tan ( angle ABC) = 15/30, and then angle ABC = 26.565051. Angle ADB is then 180 - 30 - 26.565051 or 123.434945. Using law of sines, I find that AD = 16.076951, which is approximately 16.1

[ВалерийЦелищев-шз]

Drop a perpendicular DE to AB. DE = X/2. AE = X√3/2.
Triangles ABC & EBD are similar => EB = X.
X√3/2 + X = 30. X=60/(√3 + 2) = 60((2 - √3).

[DDX]

I simply calculate the angle of B by inc tan(15/30) which is oprox=26.55 deg
now calculate the angle D as
(180-30-26.55) oprox=123.43 deg
now sin rule
(30/sin123)=AD/sin26.55
AD=(30/sin123)*sin26.55
oprox=16.08 unit

[montynorth]

Tan angle ABC = 15/30 = 0.5.
Therefore angle ABC = 26.565 degrees.
Angle ADB = 180 - 26.565 - 30 = 123.435.
Sine Rule.
AD /sin 26.565 = 30 /sin 123.435.
AD /0.4472 = 30 / 0.8345.
AD = 30 x 0.4472 / 0.8345.
AD = 16.08.

[miguelgnievesl]

From point D I drew a segment perpendicular to AB and marked it with the letter E, which generated the triangles ADE and BDE. Triangles ABC and BDE are similar, therefore their sides are in the same ratio (1:2). In the triangle BDE the side DE=a and BE=2a. So AE=30-2a. We focus on the triangle ADE. Tan30° =1/√3 = DE/AE=a/(30-2a). We solve and obtain a=30/(2+v3) or DE=30(2-v3). On the other hand since Sin30°=1/2=DE/AD then AD=60(2-v3)

[ybodoN]

Very nice trick! Here is another (longer) way to obtain an exact result: let ∠BAD = α, ∠ABD = β and ∠ADB = ɣ. Let AD = x
We know that sin α = sin 30° = 1/2. We know that tan β = 15/30 = 1/2. Using the tan to sin formula, we obtain sin β = √(1/5).
Using the sin to cos formula, we find that cos α = √(3/4) and cos β = √(4/5). Then we can calculate sin ɣ = (2√5 + √15) / 10.
Using the law of sines, we then have x = 30 √(1/5) / ((2√5 + √15) / 10). We simplify and rationalize to obtain AD = 60 (2 − √3).

[zsoltszigeti]

BD line is y=15-x/2, AD line is y*cos 30=x*sin 30 => y*√3/2=x/2. The intersection (D point) is:
x/2=15-y
y*√3/2=15-y
15=(1+√3/2)*y
y=15/(1+√3/2)
x=(15-15/(1+√3/2))*2 => x≈13.92, y≈8.04.
Length of AD is √(x^2+y^2)≈16.08

[harikatragadda]

Drop a perpendicular DE to AB.
If AD = X,
then DE = X/2 and EB = 2*DE = X
AE= 30-EB= 30-X
Applying Pythagoras theorem
X² = (30-X)² +(X/2)²
AD = X= 120 - 60√3

[waheisel]

I also created a perpendicular up from B and extended AD to that perpendicular. The length of that perpendicular segment is 30/sqrt3. You can then use the Cross Ladder Theorem to get 1/15+sqrt3/30=1/h where h is the perpendicular from AB up to D. h=30/(2+sqrt3) and AB is twice h (also 60[2-sqrt3]). Thanks PreMath!

[murdock]

Nice! ∆ ABC → AC = 15; AB = 30 →
BC = 15√5; AB = AE + BE ↔sin⁡(AED) = 1; DAE = φ = 30° → sin⁡(φ) = 1/2
ABC = δ → sin⁡(δ) = √5/5; DE ∶= a → AE = a√3 → BE = AD = 2a →
30 = a(2 + √3) → a = 30(2 - √3) → 2a = AD = 60(2 - √3)

[mathsandsciencechannel]

Nice question with great solution from premath. Am about to premiere guys. Lets go

[ritamkundu]

I really think why I couldn't find this channel before while studying in my school.

[johnlee]

Best way to remember sine and cosine of 30 45 60 degrees are
1. Write down 1 2 3 then sqrt(1) Sqrt(2) sqrt (3) then divided all by 2.
30 45 60 correspondence to 1/2 sqrt (2)/2 and sqrt(3)/2 for sine and for cosines read them as 60 45 30

[EnnioPiovesan]

We can draw from D the parallel to AB until it meets the side AC at K and the perpendicular to AB until it meets AB itself at H. Triangles ADH and ADK are both the half of an equilateral triangle, so DH=(1/2)AD and DK=(AD/2)√3.
Area(ADB) = (1/2)AB*DH=(1/2)*30*(1/2)AD
Area(ACD) = (1/2)AC*DK=(1/2)*15*(AD/2)√3
Now following your same idea:
Area(ABC)=Area(ADB)+Area(ACD) -->
15*30/2= (1/2)*30*(1/2)AD + (1/2)*15*(AD/2)√3 --> 15*15=AD[30/4+(15/4)√3 ] --> AD=15*15/[(15(2+√3))/4] = 60/(2+√3)

[wackojacko]

The 30 60 90 triangle is so cool ! ...Because the sides are in the ratio x:x⅓:2x. ...the shorter side is always x...the longer side is always x×x⅓ ...and the hypotenuse is always 2x! 🙂

[vara]

Nice problem. Please include problems on Calculus, especially limits, Integration, etc..

[melihpuskulcu]

It is also possible to solve it using law of sines. Easy to find all angles in small triangle. Then apply the rule to get the very same result.

[JSSTyger]

By dropping a line from point D perpendicular to the horizontal, we can use similar triangles. The vertical from point D has length y and the base of the newly constructed smaller triangle is, by similar triangles, equal to 2y. The overall triangle has a base of 30 but (30-2y) is the length from point A to the vertical that we just dropped from point D. We know that sin(30) = 1/2 and so the hypotenuse AD is 2y. We then complete the problem by using the pythagorean theorem and saying (2y)² = (30-2y)²+y² and solving for 2y. My answer is 16.08.