A Very Nice Geometry Problem | 2 Different Methods to Solve

A Very Nice Geometry Problem | 2 Different Methods to Solve
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[@josefriasmacedomota7884]

a = 16 . Sin x b = 16 . Cos x Área = a . b /2 = 32 16 . Sin x . 16 . Cos x /2 = 32 256 . Sin x . Cos x/2 = 32 128 . Sin x . Cos x = 32 2 . Sin x . Cos x = 32/64 Sin 2x = 1/2 2x = 30 degres x = 15 degres y = 90 - 15 = 75 degres

[@ichdu6710]

at minute 4:32 you have an equation system of two parameters. The solution: b =a/64 can be replaces in the first euqation the intoduction of the quadratic form some seconds later makes the calculation even more trickier. and no one understands why it appears here.

[@Usershima-c4v]

S=(AH×BC)/2 Then AH=4 Then AH^2=X(16-X) : 16=x(16-x) Then X=4(2+√3) tanB=AH/BH=2-√3 tanB=tan75° In this method, we do not need to get AB and AC.

[@ARN48411]

Sir, Can you make a detailed video on Chinese remainder theorem (CRT) from basics... 🙏

[@AndreasPfizenmaier-y7w]

Draw a perpendicular from A to base BC, lets call Point F. Perpendicular has got height of 4, base Segment is divided into x and 16-x. Hence 4^2=(16-x)*x. Two solutions x=14,92 and x=1,075 ( correct) . So tan Beta is 4:14,92 , Hence Beta is 15 degrees, Gamma is 90-15=75 degrees.

[@phungcanhngo]

2nd method is a lot better and easier.Thank you professor.

[@jimlocke9320]

A good educated guess is that the triangle is a special right triangle 30°-60°-90° or 45°-45°-90°. We try 30°-60°-90°, which has ratio of sides 1:√3:2, so, with hypotenuse 16, the sides are 8 and 8√3. The area is (1/2)(8)(8√3) which is 32√3 and not what we are looking for. So, we try 45°-45°-90° which has ratio of sides √2:√2:2, so, again with hypotenuse 16, the sides are 8√2 and 8√2 and area (1/2)(8√2)(8√2) = 64 and, again, not what we are looking for. However, the 15°-75°-90° triangle, while not designated "special" in geometry, appears quite often in problems, so we should make a "special" effort to memorize its properties. Its ratio of sides is (√3 - 1):(√3 + 1):2√2. So, with hypotenuse 16, its sides are (4√2)(√3 - 1) and (4√2)(√3 + 1). Its area is (1/2)(4√2)(√3 - 1)(4√2)(√3 + 1) = 16(√3 - 1)(√3 + 1) = 16(3 - 1) = 32, which is the desired area. So, the angles opposite the sides are 15° and 75°. The smaller angle must be

[@tituschirila7650]

I wasn't able to follow all 17 minutes, but height in triangle is 4 and OA is radius of the circumscribed circle, meaning is 8 (half of diameter 16 BC) => sin(angAOC) = 4/8 =1/2 angAOC is 30deg. and angABC is on circle => ang ABC = 15deg - that simple

[@aminishag8740]

Easy solution Area = 1/2 ab =32 ab = 64 -------- 1 Angle ABC = x, Angle acb = 90-x a/sinx = b/sin(90-x) = 16/sin90 = 16/1 Sin90=1 a/sinx = 16. a= 16 sinx sin(90-x) = cosx. b = 16 cost From eq 1 ab = 64 16 sinx . 16 cos x = 64 4 sinx cosx= 1 (divid by 2) 2sinx cosx = 1/2 sin2x = 1/2. = sin30 2x = 30. X= 15, 90 - x = 75

[@soli9mana-soli4953]

If there Is a thing that I've learnt about this kind of problem, it Is that Is Always convenient to find 2 theta instead of theta, computation Is much simpler😊

[@quigonkenny]

Inscribe ∆CAB in a circle. If A, B, and C are all on the circumference of a circle and ∠A = 90°, then BC is the diameter of the circle. Let O be the midpoint of BC, or the center of the circle. Draw radius OA, which will be 16/2 = 8. Draw DA, such that D is a point on OC and DA is perpendicular to BC. Let DA = h. A = bh/2 32 = 16h/2 = 8h h = 32/8 = 4 Let ∠AOD = θ. sin(θ) = DA/OA = 4/8 = 1/2 θ = sin⁻¹(1/2) = 30° Observe triangle ∆BOA. OB and OA are both radii of the circle, so OB = OA = 8. Thus ∆BOA is an isosceles triangle and ∠B = ∠OAB. If ∠AOC = θ, then ∠BOA = 180°- θ = 180 - 30 = 150°. Since ∠B = ∠OAB: ∠B = (180°- ∠BOA)/2 ∠B = (180-150)/2 ∠B = 30°/2 = 15° ∠C = 90°- 15° = 75°

[@aromaclinic4112]

EA=4 EB=x EC=16-x EB:EA=EA:EC x:4=4:(16-x) x=8±4√3 EB=8+4√3 EC=8-4√3 tan∠B=EA/EB=4/(8+4√3)=1/(2+√3） ∠B＝15゜ ∠C=75゜

[@manojkantsamal4945]

*=read as square root ^=read as to the power According to the question area of ABC =32 1/2(AB×AC)=32 AB×AC=32×2=64 Let AB=P and AC=b, So P. b=64 P^2+b^2=16^2=256=h^2 P+b=*(p^2+b^2+2pb) =*(256+2.64) =*(256+128)=*384 =8.*6.......EQN1 P-b=*(p^2+b^2-2pb) =*(256-2.64) =*(256-128) = *(128)=8.*2...Eqn2 Now add Eqn1 &Eqn2 P+b+p-b=8.*6+8.*2=8.*2(*3+1) P=4.*2(*3+1) Again substract Eqn1 &Eqn2 P+b-p+b=8.*6-8.*2 2b=8.*2(*3-1) b=4.*2(*3-1) SinC=p/h={4.*2(*3+1)}/16 =(*3+1)/2.*2 Sin(30+45)= Sin30.cos45+cos30.sin45 ={(1/2).(1/*2)}+{(*3/2).(1/*2)} =(1/2.*2)+(*3/2.*2) =(*3+1)/2.*2)=sin75 So sinc=sin75 C=75 degree B=15 degree P+b=*

[@rocksalt636]

I thought the thumbnail implied both angles were the same, in which case ? = 45 deg

[@prossvay8744]

LABC=15° ; LACB=75°

[@alexniklas8777]

15° and 75°

[@manojkantsamal4945]

75 and 15

[@축복이-x6u]

ASNWER=25 C=56

[@jan-willemreens9010]

... Good day, Area(ABC) = 32 ... Let I AB I = S & I AC I = T ... S > T ... Eq(1). S^2 + T^2 = 16^2 = 256 & Eq.(2) S * T = 2 * 32 = 64 or 4 * S * T = 256 ... so, S^2 + T^2 = 4ST .... S^2 - 4ST + T^2 = 0 .... (S - 2T)^2 - 4T^2 + T^2 = 0 ... (S - 2T)^2 = 3T^2 .... S - 2T = +/- SQRT(3)*T ... S1,2 = (2 +/- SQRT(3))*T ... recalling S > T ... so only S = (2 + SQRT(3))*T is valid .... recalling S*T = 64 ... T^2 = 64*(2 - SQRT(3)) ... T = 8*SQRT(2 - SQRT(3)) > 0 .... so, S = 8*(2 + SQRT(3))*SQRT(2 - SQRT(3)) ... finally the measure - Angle(B) = ARCTAN( T/S ) = ARCTAN( 1/(2 + SQRT(3) ) = ARCTAN( 2 - SQRT(3) ) = 15 deg. .... last but not least the measure of complementary Angle(C) of angle(B) = 90 deg. - 15 deg. = 75 deg. .... Angle(B) = 15 deg. & Angle(C) = 75 deg. .... thank you for your 2 different strategies, both of which I appreciate .... best regards, Jan-W