Find the length X | A Very Nice Geometry Problem | 2 Methods

Find the length X | A Very Nice Geometry Problem | 2 Methods

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[jimlocke]

"2nd Method" can be simplified as follows: Drop a perpendicular from D to AC and label the intersection as point E. Note that, because ΔACD is isosceles, side AC is bisected, and AE = CE = AC/2 = x/2. Consider ΔADE. Note that cos(10°) = AE/AD = (x/2)/AD and, therefore, AD = x/(2 cos(10°)). From ΔABD, at 10:12, we have AD = (2 sin(60°))/sin(100°). Later in the video, sin(100°) is replaced by equivalent cos(10°), so, doing the substitution here, AD = (2 sin(60°))/cos(10°). Equate the 2 values of AD: x/(2 cos(10°)) = (2 sin(60°))/cos(10°). Multiply both sides by 2 cos(10°) and simplify: x = 4 sin(60°). Replacing sin(60°) with (√3))/2, we get x = 2√3, as Math Booster also found.

[RAG]

Oh dear! I think both methods are excessively complicated. Having established ADC is isosceles, AD = CD. I found and expression for AD in triangle ABD using Sine rule, AD = rt3/sin100 = rt3/sin80. Then in PCD sin 80 = .5x/CD, giving CD = AD = x/2sin80. Thus rt3/sin80 = x/2sin80, so x = 2rt3.

[Monsieur_Cauchemar-YT]

As ∠BAC = 110° it is an obtuse angle it can be inscribed in a circle of radius r with centre O.
Construct lines OB, OA and OC such that they all have the same length r.
As ∠ACB = 10°, ∴ ∠AOB = 20° (angles subtended by the same arc).

Now, take △OAB:
As △OAB has lengths OB = OA = r, it is an isosceles triangle, and ∴ ∠OBA = ∠OAB = 80°.

Construct the radius from O through D which intersects line AC at D'. As both △OAC and △ADC are isosceles, lines OD' and DD' are co-linear and intersect line AC at 90°, and bisect line AC (radius-chord theorem) such that AD' = D'C = x/2.

Now, take △OAD':
As ∠OAB = 80° and ∠BAD = 100°, ∴ ∠OAD = 20°.
Now, as ∠DAC = 10° and as ∠OD'A = 90° ⇒ ∠OAC = ∠OAD + ∠DAC = 20° + 10° = 30°, ∴ ∠AOD' = 60° (angles of a triangle).

Now, take △OAD and △ABD.
∠ADB = ∠OAD = 20° and ∠AOD = ∠ABD = 60° and line AD is common to both such that △OAD ≅ △ABD (side-angle-angle).
∴ BD = OA = 2. As OA is the radius, ⇒ r = 2.

Taking △OAD' again:
As AD' = D'C = x/2 and ∠OAD + ∠DAC = 20° + 10° = 30°
cos 30° = (x/2)/2 = x/4.
⇒ x/4 = √ 3/2.
∴ x = 2√ 3.

[ahdmlk]

Sir, The complete known detail is as: Four sided shape ABCD . BD= 40 karam (1 karam =5.5 ft), AE= 16 karam ( Perpenduclar on BD). CF =25 karam (Perpendicular on BD). Area of ABCD = 820 sq karam approximately. Formula used to calculating Area is = 40(16+25)/2 . Can we find all the sides of ABCD. Thanks.

[quigonkenny]

As ∠BDA is an exterior angle to ∆ADC at D, ∠BDA = ∠CAD+∠DCA = 10°+10° = 20°. As ∠CAD = ∠DCA = 10°, ∆ADC is an isosceles triangle and AD = DC.

Method 1:
Drop a perpendicular from D up to CA at E. As ∆ADC is an isosceles triangle, by rule, DE bisects ∆ADC and forms two congruent right triangles ∆CED and ∆DEA. As ∠EAD = ∠DCE = 10°, ∠ADE = ∠EDC = 90°-10° = 80°.

As ∠ABD = 60° and ∠BDA = 20°, ∠DAB = 180°-(60°+30°) = 100°. Extend BA to F and draw DF such that FA = PD, ∠DFA = 90°, and, as ∠FAD = 180°-100° = 80°, ∠ADF = 90°-80° = 10°. As DA is common, by construction, ∆DFA is congruent with ∆DEA (and by extension ∆CED.

As ∠ADF = 10° and ∠BDA = 20°, ∠BDF = 30° and ∆BDF is a 30-60-90 special right triangle, such that FB = BD/2 = 1 and DF = √3FB = √3.

As FD = √3 and CE = EA = FD, then x = CA = CE + EA = √3 + √3 = 2√3.

x = 2√3 units.

Method 2:
As ∠BDC is a straight angle, ∠ADC = 180°-20° = 160°. As ∠ADC is an exterior angle to ∆BDA at D, ∠DAB = ∠ADC-∠ABD = 160°-60° = 100°.

By the law of sines:

BD/sin(100°) = DA/sin(60°)
DA = BDsin(60°)/sin(100°)
DA = 2(√3/2)/sin(100°) = √3/sin(100°)
DC = DA = √3/sin(100°)
DC = √3/cos(10°)

DC/sin(10°) = CA/sin(160°)
CA = DCsin(160°)/sin(10°)
x =
x = √3sin(20°)/sin(10°)cos(10°)
x =
x = 2√3 units

[timeonly]

Love 1st method! ❤❤
Over and over, we see that good & clever geometry > trigonometry!!
Now, if only I can see the geometry!!! 😅

[alokranjan]

Can be easily solved by using both cosine & sine formula in a triangle

[幕天席地-wc]

Let O be the centre of the circumscribed circle of triangle ABC. Angle AOB=20, AOC=120, so OBD=20. Traingle AOD and COD are congruent, so angle AOD=60, so angle BOD=BDO=80, so BO=BD=2. Angle OAC=30, so AC=2*sqrt(3).

[manojitmaity]

Sir upload some algebra videos also, please.

[holyshit]

ADC is isosceles so |AD|=|DC|
Measure of angle ADC = 160 degrees
Measure of angle ADB = 20 degrees
Measure of angle BAD = 100 degrees


From sine law in ABD

2/sin(100) = |AD|/sin(60)
|AD| = 2sin(60)/sin(100)
|AD| = sqrt(3)/sin(80)
|AD| = sqrt(3)/cos(10)

From cosine law in ADC

x^2
x^2 = 6/cos^2(10) +6/cos^2(10)*cos(20)
x^2 = 6/cos^2(10)*(1+cos(20))
x^2 = + cos^2(10) - sin^2(10))
x^2 = 6/cos^2(10)*(2*cos^2(10))
x^2=12
x = 2*sqrt(3)