First impression : very difficult😢.

@PreMath

10 ай бұрын

Thanks for your feedback! Cheers! 😀

Once we have found BC = 6 and AB = BE = 10, we also know that CE = 8. So x = arctan 10/6 − arctan 8/6 ≈ 5.91° If we use use trigonometric identities to get an exact value: tan x = (10/6 − 8/6) / (1 + 10/6 ⋅ 8/6) ⇒ tan x = 3/29

Without using area of blue triangle...we can also find angle x by using the law of cosine. Thanks for your problem. I always follow your newest math problem. Warm regards from Indonesia.

TanX = Tan(∠DBC - ∠EBC) = (DC/BC - EC/BC)/{1+(DC*EC/BC*BC)}= 3/29

I got 5.906 degrees: Invoking Pythagoras, I got the dimensions of the rectangle as 6x10 by a slight variation on your method; but I'll skip the details. Then angle x = angle ABE -- angle ABD. Tan ABE = 6/8 = 0.75; and tan ABD = 6/10 = 0.60. So angle x = arctan(0.75) -- arctan(0.60) = 36.8699 -- 30.9637 = 5.906 degrees. Cheers. 🤠

We note that

U have very kind community, pre math. I love it =)

@PreMath

10 ай бұрын

I'm very blessed. So nice of you. Thank you! Cheers! 😀

Impressed, all you need is just to focus a little bit.

Good explanation. I did the final part with

@PreMath

10 ай бұрын

Thank you! Cheers! 😀

Nice and awesome, many thanks! AB = BE = y area ∆BED = 6 = gh/2 = 2h/2 = h = AD → sin⁡(δ) = (y - 2)/y → cos⁡(δ) = 6/y → sin^2(δ) + cos^2(δ) = 1 → 4y = 40 → y = 10 → y - 2 = 8 → BD = √136 = 2√34 → 6 = (1/2)10(2√34)sin⁡(φ) → sin⁡(φ) = 3√34/170 → φ ≈ 5,9° or: tan⁡(θ) = 3/5; tan⁡(τ) = cot⁡(θ) = 5/3; tan⁡(δ) = 4/3 → tan⁡(τ - δ) = (tan⁡(τ) - tan⁡(δ))/(1 + tan⁡(τ)tan⁡(δ)) = 3/29 → τ - δ = φ ≈ 5,9°

@PreMath

10 ай бұрын

Excellent! Thank you! Cheers! 😀

Keep up the great page! I am number one fun!

@PreMath

10 ай бұрын

Thanks! Will do! Take care dear

What is the Gaussian Radius of these square units? Just asking for a friend......never mind, I answered my own question . Constant Zero Gaussian surfaces!.... and that's when the study of Euclidean geometry becomes so much fun with you Sir. 🙂

Thanks for video.Good luck sir!!!!!!!

@PreMath

10 ай бұрын

Thank you too

Very well explained 👍 Thanks for sharing😊

@PreMath

10 ай бұрын

Thanks for liking

Yeah, maybe I'm not on form or just plain dumb, didn't manage this on but still got much enjoyment from it. Thank you.

@PreMath

10 ай бұрын

You are very welcome! Thank you! Cheers! 😀

Nice sir🎉

Awesome question 😅

Using Trig. as alternative once length y was calculated. Tan EBC = 8/6, EBC = 53.13 degrees. Tan ABD = 6/10, ABD = 30.96 degrees. Then x = 90 - 53.13 - 30.96 = 5.91 degrees.

6:30 PM in HK, very early in USA 😮

@PreMath

10 ай бұрын

Very true! It's 3:30 AM here in the USA. Take care. Thank you! Cheers! 😀

0.5*AD*DE = 6. AD = 6. Let AB = A. Asq = 36 + (A-2)sq. A = 10. Angle X = InvTan (6/8) - InvTan (6/10) = 5.9 Deg.

thankyou so much dear ❤❤

@PreMath

10 ай бұрын

You are so welcome! So nice of you, dear Thank you! Cheers! 😀

Yay! I solved it.

@PreMath

10 ай бұрын

Awesome!

I find it quite ez… the key here is to find the length of AD, then just use equation… very ezzzz…

Area of blue triangle with b=2 : A = ½ b. h = ½ 2 h = 6 cm² h = 6 cm Side EB = base of rectangle b² = (b-2)²+h² b² = (b² -4b + 4) + h² 4b = h² + 4 b = h²/4 + 1 = 6²/4+1 b = 10 cm x = α - β tan α = 10/6 , α = 59,04° tan β = 8/6 , β = 53,13° x = 5,906° ( Solved √ )

Why is the answer wrong when i use trig ratios to get the side length DB?

x=arctg3/29

By the foolish way, let axb be the dimension of the rectangle, so BD^2=a^2+b^2, BE^2=b^2=a^2+(b-2)^2, so a^2=4b-4, and 144=(a^2+b^2)b^2sin^2 x=(b^2+4b-4)b^2sin^2 x, 4=b^2+(b^2+4b-4)-2b sqrt(b^2+4b-4)cos x, it becomes very clumsy to solve for x and b from these two equations 😅

Tan-1(6÷10)-Tan-1(8÷10)

Good night sir

@PreMath

10 ай бұрын

Same to you, dear

as n were=25 os is it

It's better to call the unknown angle alpha.

@PreMath

10 ай бұрын

Thank you! Cheers! 😀

@ybodoN

10 ай бұрын

I use alpha for a first angle (followed by beta, gamma, delta). Where there is only one unknown angle, I prefer to use theta. But the Greek letter chi (Χ or χ) could also be a valid option...

Why DE is a base of triangle not DB? and height triangle DEB why took DA...DA is height of triangle DAB

@thewolfdoctor761

10 ай бұрын

Because initially we know the value of DE to be 2 but we don't know the value of DB, while the area is given as 6.

❤🥂👍😀

@PreMath

10 ай бұрын

So nice of you. Thank you! Cheers! 😀

so difficult 🥲🥲🥲🥲

Me 5,91°

Area of blue triangle : b=2cm A = ½ b. h = 6 cm² A = ½ 2 h h = 6 cm Side EB = base of rectangle b² = (b-2)²+h² b² = (b² -4b + 4) + h² 4b = h² + 4 b = h²/4 + 1 = 6²/4+1 b = 10 cm Area of blue triangle: A = ½ b.d sinx sin x = 2A/(b.d) sin x = 2A/(b.√(b²+h²)) sin x = 2.6/(10√(10²+6²)) sin x = 12 /(10.11,66) sin x = 0,1029 x = 5,906° ( Solved √ )