A Very Nice Geometry Problem | 2 Different Methods to Solve
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- Опубліковано 16 бер 2024
- A Very Nice Geometry Problem | 2 Different Methods to Solve
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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
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.
Sir, Can you make a detailed video on Chinese remainder theorem (CRT) from basics... 🙏
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
2nd method is a lot better and easier.Thank you professor.
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.
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
LABC=15° ; LACB=75°
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😊
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°
*=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=*
I thought the thumbnail implied both angles were the same, in which case ? = 45 deg
15° and 75°
75 and 15
ASNWER=25 C=56
... 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