The 15°-75°-90° right triangle is not considered "special" in geometry, but appears frequently in problems. Knowing that its long side is 2 + √3 times as long as its short side provides a shortcut to this problem's solution. At 0:50, the vertices of the triangles have been given designations. Drop a perpendicular from R to PS and label the intersection as point T. ΔRST is an isosceles right triangle with hypotenuse RS = a, so sides RT and ST have length a/(√2). Consider ΔPRT, which is a 15°-75°-90° right triangle. Its long side PT is 2 + √3 times as long as its short side RT, so PT has length (2 + √3)a/(√2). Drop a perpendicular from Q to PS and label the intersection as point U. ΔQSU is an isosceles right triangle with hypotenuse QS = 3a, so sides QU and SU have length 3a/(√2). Length TU = SU - ST = 2a/(√2). Length UP = PT - TU = (2 + √3)a/(√2) - 2a/(√2) = (√3)a/(√2). Consider right ΔPQU. The ratio of sides QU/UP = (3a/(√2))/((√3)a/(√2)) = (3a)/((√3)a) = √3. So side QU is √3 times as long as side UP. This is the ratio long side to short side for a 30°-60°-90° right triangle. So,
Yes sir 🤗, when I started reading, you said "the 15°-75°-90° right triangle is not considered a "special" in geometry" I started asking myself, did make a mistake by saying in the video that it was considered "special"🤣 But when I continued reading it, I realized you were trying to prove a point. yes friend this is a perfect solution, thanks
@@MathandEngineering Thanks for the compliment! In the video, I find no mention of the 15°-75°-90° right triangle but the 30°-60°-90° and 45°-45°-90° "special" right triangles appear. At 4:05, we can reach the solution faster by finding that ΔQAP is a 15°-75°-90° right triangle with
(a=sqrt(2) because i'm worth it ; then a*cos(45°)=a*sin(45°)=1) X=45°+arctan((a*cos(45°)+a*sin(45°)*tan(90°-15°)-3*a*cos(45°))/(3*a*sin(45°))) X=45°+arctan((1+1*tan(90°-15°)-3)/3) X=45°+arctan((1+(2+sqrt(3))-3)/3) X=45°+arctan(sqrt(3)/3) X=45°+30° X=75°
I think that once you found that QS = 3√ 3 - 3 you can easily find QA that is QA = 3√ 3 - 3 - 1 - √ 3 + 1 = 2√ 3 - 3 and knowing that AP = √ 3 you can find tan Q = √ 3/(2√ 3 - 3) = 2 + √ 3 that means the angle in Q is 75°
At 1:32, "We'll assume the length of AR to be 1". Pardon? It is actually a(rt3+1)/2. And z is a(3+rt3)/2 later. QA is a(3-rt3)/2, so tan x is 2+rt3 which means it is 75 degrees, but your method is nonsense.
He is scaling, assuming 1 is okay because "a" is a variable affecting only one side of the triangle. So you can scale the triangle without affecting the angles. I used the Law of Sines, and the result is the same.
The 15°-75°-90° right triangle is not considered "special" in geometry, but appears frequently in problems. Knowing that its long side is 2 + √3 times as long as its short side provides a shortcut to this problem's solution. At 0:50, the vertices of the triangles have been given designations. Drop a perpendicular from R to PS and label the intersection as point T. ΔRST is an isosceles right triangle with hypotenuse RS = a, so sides RT and ST have length a/(√2). Consider ΔPRT, which is a 15°-75°-90° right triangle. Its long side PT is 2 + √3 times as long as its short side RT, so PT has length (2 + √3)a/(√2). Drop a perpendicular from Q to PS and label the intersection as point U. ΔQSU is an isosceles right triangle with hypotenuse QS = 3a, so sides QU and SU have length 3a/(√2). Length TU = SU - ST = 2a/(√2). Length UP = PT - TU = (2 + √3)a/(√2) - 2a/(√2) = (√3)a/(√2). Consider right ΔPQU. The ratio of sides QU/UP = (3a/(√2))/((√3)a/(√2)) = (3a)/((√3)a) = √3. So side QU is √3 times as long as side UP. This is the ratio long side to short side for a 30°-60°-90° right triangle. So,
Yes sir 🤗, when I started reading, you said "the 15°-75°-90° right triangle is not considered a "special" in geometry"
I started asking myself, did make a mistake by saying in the video that it was considered "special"🤣
But when I continued reading it, I realized you were trying to prove a point. yes friend this is a perfect solution, thanks
@@MathandEngineering Thanks for the compliment! In the video, I find no mention of the 15°-75°-90° right triangle but the 30°-60°-90° and 45°-45°-90° "special" right triangles appear.
At 4:05, we can reach the solution faster by finding that ΔQAP is a 15°-75°-90° right triangle with
(a=sqrt(2) because i'm worth it ; then a*cos(45°)=a*sin(45°)=1)
X=45°+arctan((a*cos(45°)+a*sin(45°)*tan(90°-15°)-3*a*cos(45°))/(3*a*sin(45°)))
X=45°+arctan((1+1*tan(90°-15°)-3)/3)
X=45°+arctan((1+(2+sqrt(3))-3)/3)
X=45°+arctan(sqrt(3)/3)
X=45°+30°
X=75°
Wow, a trigonometric approach, it's excellent sir, thanks
I think that once you found that QS = 3√ 3 - 3 you can easily find QA that is
QA = 3√ 3 - 3 - 1 - √ 3 + 1 = 2√ 3 - 3
and knowing that AP = √ 3
you can find tan Q = √ 3/(2√ 3 - 3) = 2 + √ 3
that means the angle in Q is 75°
Why AR = 1? It's not necessary.
φ = 30°; QO = PO = SO = r; PSQ = 3φ/2; PR = PO + RO → sin(QOR) = 1
OPM = MSO = φ/2 → SQO = OSQ = φ → QS = QN + SN = QS/2 + QS/2 = 3a →
NOR = 2φ = NOR + ROS = φ + φ; QO = PO = r → OQP = QPO = θ →
2θ = 6φ - (OPS + PSQ + SQO) = 6φ - (φ/2 + 3φ/2 + φ) = 3φ →
θ = 3φ/2 → SQP = 3φ/2 + φ = 5φ/2
or: PSQ = 3φ/2 → POQ = 3φ → QO = PO = r →
QPO = POQ = 3φ/2 → SQP = (φ/2)(3 + 2)
{2a+2a ➖ }{15°A+45°B+90°C}=4a^2{150°ABC+30°D}=180°ABCD/4a^2=40a^4.20ABCD 2^20a^2^2.2^10 1^10a^1^1.2^2^5 2^5.1^1^1 2^1.1 2.1 (ABCD a ➖ 2ABCD a+1).
Why did you asume that AR=1????
At 1:32, "We'll assume the length of AR to be 1". Pardon? It is actually a(rt3+1)/2.
And z is a(3+rt3)/2 later. QA is a(3-rt3)/2, so tan x is 2+rt3 which means it is 75 degrees, but your method is nonsense.
He is scaling, assuming 1 is okay because "a" is a variable affecting only one side of the triangle. So you can scale the triangle without affecting the angles. I used the Law of Sines, and the result is the same.
2a/sin(120-x)=t/sinx...a/sin15=t/sin45...divido le equazioni,rimane un'equazione in x..ctgx=((4sin15/sin45)-1)/√3..tgx=2+√3...x=75