Can you find angle X? | Area of the Blue Shaded triangle is 6 | Important Geometry skills explained
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- Опубликовано: 29 сен 2024
- Learn how to find the angle X in the blue shaded triangle. Important Geometry skills are also explained: area of the triangle formula; Trigonometry; Pythagorean theorem. Step-by-step tutorial by PreMath.com
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Can you find angle X? | Area of the Blue Shaded triangle is 6 | Important Geometry skills explained
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First impression : very difficult😢.
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
Awesome question 😅
Tan-1(6÷10)-Tan-1(8÷10)
❤🥂👍😀
So nice of you.
Thank you! Cheers! 😀
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. 🤠
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.
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 √ )
We note that
Why DE is a base of triangle not DB? and height triangle DEB why took DA...DA is height of triangle DAB
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.
U have very kind community, pre math. I love it =)
I'm very blessed.
So nice of you.
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°
Excellent!
Thank you! Cheers! 😀
Good explanation. I did the final part with
Thank you! Cheers! 😀
Quicker Solution at the end: 6 and 10 are my lengths for tan BETA , 6 and 8 are side lengths for tan E , angle sum in the two triangles 180 , 90 degrees- 53,1 degrees- 30,96 degrees = 5,93 degrees
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. 🙂
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 😅
x=arctg3/29
Nice sir🎉
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.
You are very welcome!
Thank you! Cheers! 😀
It's better to call the unknown angle alpha.
Thank you! Cheers! 😀
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...
Thanks for video.Good luck sir!!!!!!!
Thank you too
I find it quite ez… the key here is to find the length of AD, then just use equation… very ezzzz…
Yay! I solved it.
Awesome!
Impressed, all you need is just to focus a little bit.
Good night sir
Same to you, dear
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?
Great explanation!
It seems like h is not part of the triangle DBE.
But why did you use h as Height in this : 6 = 1/2(2)(h) ?
Mr African, good question. I was also puzzled. But if you look again at the blue triangle the base is 2. Now think again about the height. It does not have to be inside the triangle. Any perpendicular distance from the opposite sides will be the height.
thankyou so much dear ❤❤
You are so welcome!
So nice of you, dear
Thank you! Cheers! 😀
Very well explained 👍
Thanks for sharing😊
Thanks for liking
as n were=25 os is it
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.
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.
6:30 PM in HK, very early in USA 😮
Very true! It's 3:30 AM here in the USA.
Take care.
Thank you! Cheers! 😀
Keep up the great page! I am number one fun!
Thanks! Will do!
Take care dear
so difficult 🥲🥲🥲🥲
Me 5,91°