You can use the sine rule. sin x / 50 = sin 2x / 80 sin x /50 = 2 sin x cos x / 80 cos x = 80 / 100 = 4/5 From which sin x = 3/5 (x is the small angle in a 3-4-5 triangle) height h = 80 sin x = 48 base b = 80 cos x + 50 cos 2x cos 2x = cos^2 x - sin^2 x = 7/25 b = 64 + 14 = 78 Area = 48 x 78 / 2 = 1872
@@ybodoN That was my first way of solving. I wrote four and two ways were lost. Please check them. I typed it first, they were lost and instead of having 10+ likes i have none!!!
@@ybodoN i was writing 1,5 hours into laptop.Isaved iit but it hasn't square root symbol.Immediately i started writing from my mobile phone,which has symbols of square root. When i finished the transformation i thought i presed send. Unfortunatelly after 1 hour and 45 minutes everything was lost. Then i was writing another 1 hour. My wife said these days i have no man. And you have right,that's life. Now i'm 60 years old and i have seen it a long ago.
Alternate method using law of sines: 50 / sin x = 80 / sin 2x sin 2x / sin x = 80 / 50 2 sin x cos x / sin x = 8/5 cos x = 4/5 sin x = 3/5 Now we find sin C sin C = sin(180−x−2x) = sin(180−3x) = sin 3x = 3 sin x − 4 sin³x = 3(3/5) − 4(27/125) = 117/125 And finally, we can calculate area of triangle Area(ABC) = 1/2 (AC)(BC) sin C = 1/2 (50) (80) (117/125) = 1872
I bisected the 2x angle then used similar triangles to get the third side of 78. Then used Heron's formula to get the 1872 area. As usual PreMath's solution was more elegant. Thanks again for the fun math puzzle!
Drop a perpendicular from point D to BC and label the intersection as point F. Note that BC is divided in half, length BF = CF = 40 and two right triangles are formed, ΔBDF and ΔCDF. Hypotenuse BD of ΔBDF has length 50. By Pythagorean theorem on ΔBDF with hypotenuse 50 and side length BF = 40, then side length DF = 30 (familiar 3-4-5 right triangle, scaled up by factor of 10.) ΔBCE and ΔBDF are similar by angle-angle (
Being 'B' the base of triangle, and 'H' the height: B = 80 cos x + 50 cos 2x also : B²= 50² + 80² - 2. 50.80 cos (180°-3x) [Cosine theorem] We have 2 formulas and 2 unknown (B and angle x) Then the problem is solved Puting theese formulas in an Excel worksheet, i have obtained: Angle x= 36.87° B = 78 cm H = 48 cm Area = B.H /2 = 1872 cm² (Solved)
My fourth geometrical way is the same as Prof.Premath until the calculation of h, by an easy way.WE DRAW DF altitude of isosceles triangle DCB.Then DF is middle bisector of CB,so CF=FB=40. Then use the Rule AREA is Standard ,so AREA of CDB=1/2*DB*CE=1/2*CB*DF so 50*h=80*sqrt(50^2-40^2) so 50*h=80*30 so h=2400/50=48. (DF^2=DB^2-FB^2). Then ED^2=50^2-h^2=50^2-48^2=2500-2304=196 so ED=sqrt(196)=14 and AD=2*14=28 and AB=AD+DB=28+50=78. Finally Area=1/2*AB*h=1/2*78*48=1872. Conclusion the Trigonometrical Method is the easiest of four methods i've shawn.
Theorem of sines: sin(x)/50 = sin(2x)/80 Trigonometric identity: sin(2x) = 2 sin(x) cos(x) sin(x)/50= 2 sin(x) cos(x) / 80 80.sin(x)= 100 sin(x) cos(x) 80 = 100 cos(x) cos(x) = 80/100 = 0.8 x= 36.8699° Being 'B' the base of triangle, and 'H' the height: B = 80 cos(x) + 50 cos(2x) B = 78 cm H = 80 sin(x) H= 48 cm Area = B.H/2 Area = 1872 cm² (Solved)
Being 'H' the height of triangle. H = 80 . sin(x) = 50 . sin(2x) sin(2x) = 2 sin(x) cos(x) 80. sin(x) = 50. 2 sin(x) cos(x) 80 = 100 cos(x) cos(x) = 80/100 = 0.8 x= 36.8699° Area of any triangle: A = ½ Side1 Side2 sin(Angle) A = ½ 50 . 80 . sin (180°-3x) A = 1872 cm² (Solved )
I did it this way, Sinx=h/80 Sin2x=h/50=2sinxcosx=2(h/80)cosx This gives cosx=4/5 Sin^2x+16/25=1 Gives sinx=3/5 Substitute back to get h=48 Using pythagoras twice gives the bases as 14 and 64 for a total 78 A=.5*78*48=1872 square units
I dare say it is not a good practice to use "x" to measure both angles and lines in one problem and even more in one triangle. ))))) x could be left for angles, and y (for instance) used for line measurements. Anycase 👍
Seeing the angles x and 2x made me think of the trigonometric approach. No peeking, no calculators: Drop the perpendicular CE. Let CE = h (height of the big triangle); AE = w; and EB = z. Then sin 2x = h/50 = 2 sin x cos x = (2)(h/80)(z/80) = hz/3200. So h/50 = hz/3200; multiply both sides by 50: h = hz/64; divide both sides by h: 1 = z/64; so z = 64. From here it's just a matter of invoking Pythagoras twice: h^2 = 80^2 - 64^2 = 6400 - 4096 = 2304; h = √2304 = √((16)(144)) = (√16)(√144) = (4)(12) = 48; and then w^2 = 50^2 - 48^2 = 2500 - 2304 = 196; w = √196 = √((4)(49)) = (√4)(√49) = (2)(7) = 14. So AB = 14 + 64 = 78; and the area is (1/2)(48)(78) = (24)(78) = 1872. And it would have been quicker if I'd realized that both right triangles are integer Pythagoreans: AEC is 14-48-50 and EBC is 3-4-5 x 16. But I didn't notice that until the end. Cheers. 🤠
La vertical por “C” corta a AB en “E”→ Tomando CE como eje, obtenemos el punto D simétrico de A → ∠ADC=2X → CA=50=CD=DB → AB=AE+ED+DB =b+b+50 → 50²-b² = CE²= 80²-(b+50)² → b=14 → AB=(2*14)+50=78 → CE²=50²-b² →CE=48 → Área ∆ABC =78*48/2=1872 Gracias y saludos.
At a quick glance, Using the sine rule Sin(x)/50 = sin(2x)/80 = sin(C)/AB and the Trig function Sin(2x)= 2*sin(x) * cos(x) Then sin(x)/sin(x)cos(x)= 100/80 then cos(x) = 80/100 and x= 36.87, 2*x = 73.74 and Angle ACB = 180-3*x =69.4. Then AB = (Sin(69.4)/sin(36.87))*50 = 78. area is half base * height = 39 * sin(36.87) * 80 = 1872 area units.
The content of your channel is exactly what I was looking for! Kudos for the good work sir! Please keep uploading more of these plane geometry problems. Very helpful.
Brilliant exercise, looks simple, and is quite simple, but the answer doesnt just jump out at you. You have to make some effort to work your way to the result and that makes it both interesting and challenging. Only humans I think can plan strategies for problem solving, its a gift we should all try to develop, often frustrating but always rewarding. Thanks again
You can use Sin2x/80 = Sinx50, and then use the identity Sin2x = 2SinxCosx, from there make Cosx the subject of the formula, it's an easier method for me
Once you determine that x = 14, there's no need to calculate to find h, as [7, 24, 25] is a primitive Pythagorean Triple, and we already have a 14 (7×2) and a 50 (25×2), meaning 48 (24×2) is the other side.
50/sinx = 80/sin2x. General formula, sin2x = 2.sinx.cosx. Therefore 50/sinx =80/2.sinx.cosx. 50.2.sinx.cosx = 80.sinx. 100 sinx.cosx = 80.sinx. Cos x = 80.sinx/100sinx. Cos x = 0.8. Cos -1, 0.8 = 36.8698 degrees. Angle ACB =180 - 3x. ACB = 180 - 110.6094. ACB =69.3906 degrees. Area of triangle = 1/2 x 50 x 80 x sin 69.3906. 1872.
Using the trigonometric identities, we can calculate that the sine of the angle ACB is very precisely 117/125😉 (this without ever going through the angle values in degrees, imprecise even when rounded to the 4th decimal)
Extending AB side on the left to point F so that CF= CB=80. (FCB is isoscele) angle CFA = x by construnction angle FCA = x because of the sum of external angle ACF triangle is also isoscele so FA = CA = 50 I found the area of ACF triangle with Erone's formula (only to use something different of Pythagora...) and got Area = 1200 then found the height of ACF , CE=(1200*2)/50 = 48 then calculated AE and EB with Pythagorean Theorem... but I prefer trigonometric solution
Draw a Perpendicular h from C to AB. Sin 2x = h/50. Sin x = h/80. h = 80*Sin x. SIn 2x = 80*Sin x /50. 2(Sin x)*(Cos x) = 1.6*Sin x. Cos x = 0.8 = EB/80. EB = 64. Cos 2x = 2Cossqx - 1 = .28 = AE/50. AE = 14. AB = 14+64 = 78. x 36.87. h = 80*Sin36.87 = 48. Area = 0.5*48*78 = 1872
Καλημέρα σας. Μπορούμε να πάρουμε το επεκτεταμένο (Γενικευμένο) Πυθαγόρειο θεώρημα για την αμβλεία γωνία BDC του ισοσκελούς τριγώνου BDC και να υπολογίσουμε το x. Μετά με Πυθαγόρειο θεώρημα στο CDΕ, υπολογίζουμε το h. ...
I agree with th solutions shown, but I disagree with the fact that AB segment is smaller than the CB segment (particularly, 74 versus 80) as seen on the drawing. Whoever came with this problem should have used real values for angles to calculate the sides, and use the real proportions between the two sides given ( AC and BC). The values have to make sense too.At least for me...
When i was typing my rule of sines method.,i tried to show you the calculation of Area by three differnt ways. I pressed something and everything was lost. Luckily MrPwmiles solved it by one of my three ways. I'll discribe the other two ways. First Trigonometrical an second geometrical.Rule of Sines gave sin (2x)=80*sinx/50 so 2*sinx*cosx=80*sinx/50 so cosx=80/100=4/5 and sinx=3/5. Formula of Area of Triangle is **1/2*a*b*sin(
Because the sum of theinterior angles of triangle CAD equals 180⁰ and angles CDA and CDB make a straight angle 180⁰...so subtract angle CDA from both interior triangle and straight angles and wallah! The sum of opposite interior angles. 🙂
So h/50 = 2(h/80)(EB/80) The h cancels out, allowing for a quick solve for EB. The rest of the triangle then just comes together using the pythagorean theorem.
Out box? First I would do it in a traditional way. Draw CD, where D on AB and angle ADC 2x, thus the triangle DBC is an isosceles triangle with sides 50-80-50, then clearly the height of this triangle is 30, and sin x is 3/5, and the angle outside ACB is 3x, the height of the great triangle is 50xsin 3x=50(3sin x-4sin^3 x)=50(9/5-108/125)=90-43.2=46.8, therefore the answer is 80x46.8/2=1872.🙂
It’s always a pleasure listening to your dulcet tones!❤🥂😊
Wow, thank you!
Take care, my dear friend. Stay blessed as always🙏
@@PreMath ❤️😀🍻
You can use the sine rule.
sin x / 50 = sin 2x / 80
sin x /50 = 2 sin x cos x / 80
cos x = 80 / 100 = 4/5
From which
sin x = 3/5 (x is the small angle in a 3-4-5 triangle)
height h = 80 sin x = 48
base b = 80 cos x + 50 cos 2x
cos 2x = cos^2 x - sin^2 x = 7/25
b = 64 + 14 = 78
Area = 48 x 78 / 2 = 1872
Thank you! Cheers! 😀
The height calculation can be ignored since the area is ½ 80 · 78 · sin x😉
@@ybodoN That was my first way of solving. I wrote four and two ways were lost. Please check them. I typed it first, they were lost and instead of having 10+ likes i have none!!!
@@sarantis40kalaitzis48 MrPwmiles successfully published his trigonometrical (easy to read) solution earlier than you... That's life...
@@ybodoN i was writing 1,5 hours into laptop.Isaved iit but it hasn't square root symbol.Immediately i started writing from my mobile phone,which has symbols of square root. When i finished the transformation i thought i presed send. Unfortunatelly after 1 hour and 45 minutes everything was lost. Then i was writing another 1 hour. My wife said these days i have no man. And you have right,that's life. Now i'm 60 years old and i have seen it a long ago.
Alternate method using law of sines:
50 / sin x = 80 / sin 2x
sin 2x / sin x = 80 / 50
2 sin x cos x / sin x = 8/5
cos x = 4/5
sin x = 3/5
Now we find sin C
sin C = sin(180−x−2x) = sin(180−3x) = sin 3x = 3 sin x − 4 sin³x = 3(3/5) − 4(27/125) = 117/125
And finally, we can calculate area of triangle
Area(ABC) = 1/2 (AC)(BC) sin C = 1/2 (50) (80) (117/125) = 1872
supported
I bisected the 2x angle then used similar triangles to get the third side of 78. Then used Heron's formula to get the 1872 area. As usual PreMath's solution was more elegant. Thanks again for the fun math puzzle!
Drop a perpendicular from point D to BC and label the intersection as point F. Note that BC is divided in half, length BF = CF = 40 and two right triangles are formed, ΔBDF and ΔCDF. Hypotenuse BD of ΔBDF has length 50. By Pythagorean theorem on ΔBDF with hypotenuse 50 and side length BF = 40, then side length DF = 30 (familiar 3-4-5 right triangle, scaled up by factor of 10.) ΔBCE and ΔBDF are similar by angle-angle (
By the Sine Rule, sin(x)/50 = sin(2x)/80 = 2.sin(x).cos(x)/80.
So cos(x) = 4/5 and sin²(x) = 1- (4/5)² = 9/25. ∴ sin(x) = 3/5.
So area(Δ ABC) = (½)|AC|.|BC|.sin(180°-3x) = (½)(50)(80)sin(3x)
=25(80)sin(x){3- 4sin²(x)} =25(80)(3/5)(3- 36/25) =16(3)(39)=1872.
Being 'B' the base of triangle, and 'H' the height:
B = 80 cos x + 50 cos 2x
also :
B²= 50² + 80² - 2. 50.80 cos (180°-3x) [Cosine theorem]
We have 2 formulas and 2 unknown (B and angle x)
Then the problem is solved
Puting theese formulas in an Excel worksheet, i have obtained:
Angle x= 36.87°
B = 78 cm
H = 48 cm
Area = B.H /2 = 1872 cm² (Solved)
My fourth geometrical way is the same as Prof.Premath until the calculation of h, by an easy way.WE DRAW DF altitude of isosceles triangle DCB.Then DF is middle bisector of CB,so CF=FB=40. Then use the Rule AREA is Standard ,so AREA of CDB=1/2*DB*CE=1/2*CB*DF so 50*h=80*sqrt(50^2-40^2) so 50*h=80*30 so h=2400/50=48. (DF^2=DB^2-FB^2). Then ED^2=50^2-h^2=50^2-48^2=2500-2304=196 so ED=sqrt(196)=14 and AD=2*14=28 and AB=AD+DB=28+50=78. Finally Area=1/2*AB*h=1/2*78*48=1872. Conclusion the Trigonometrical Method is the easiest of four methods i've shawn.
Theorem of sines:
sin(x)/50 = sin(2x)/80
Trigonometric identity:
sin(2x) = 2 sin(x) cos(x)
sin(x)/50= 2 sin(x) cos(x) / 80
80.sin(x)= 100 sin(x) cos(x)
80 = 100 cos(x)
cos(x) = 80/100 = 0.8
x= 36.8699°
Being 'B' the base of triangle, and 'H' the height:
B = 80 cos(x) + 50 cos(2x)
B = 78 cm
H = 80 sin(x)
H= 48 cm
Area = B.H/2
Area = 1872 cm² (Solved)
Being 'H' the height of triangle.
H = 80 . sin(x) = 50 . sin(2x)
sin(2x) = 2 sin(x) cos(x)
80. sin(x) = 50. 2 sin(x) cos(x)
80 = 100 cos(x)
cos(x) = 80/100 = 0.8
x= 36.8699°
Area of any triangle:
A = ½ Side1 Side2 sin(Angle)
A = ½ 50 . 80 . sin (180°-3x)
A = 1872 cm² (Solved )
I did it this way,
Sinx=h/80
Sin2x=h/50=2sinxcosx=2(h/80)cosx
This gives cosx=4/5
Sin^2x+16/25=1
Gives sinx=3/5
Substitute back to get h=48
Using pythagoras twice gives the bases as 14 and 64 for a total 78
A=.5*78*48=1872 square units
I dare say it is not a good practice to use "x" to measure both angles and lines in one problem and even more in one triangle. )))))
x could be left for angles, and y (for instance) used for line measurements.
Anycase 👍
Even better: uppercase Latin letters for vertices, lowercase Latin letters for sides and lower case Greek letters for angles.
@@ybodoN 👍
That's the way I've learned it at school, back then...
Seeing the angles x and 2x made me think of the trigonometric approach. No peeking, no calculators:
Drop the perpendicular CE. Let CE = h (height of the big triangle); AE = w; and EB = z. Then
sin 2x = h/50 = 2 sin x cos x = (2)(h/80)(z/80) = hz/3200. So
h/50 = hz/3200; multiply both sides by 50:
h = hz/64; divide both sides by h:
1 = z/64; so z = 64. From here it's just a matter of invoking Pythagoras twice:
h^2 = 80^2 - 64^2 = 6400 - 4096 = 2304;
h = √2304 = √((16)(144)) = (√16)(√144) = (4)(12) = 48; and then
w^2 = 50^2 - 48^2 = 2500 - 2304 = 196;
w = √196 = √((4)(49)) = (√4)(√49) = (2)(7) = 14.
So AB = 14 + 64 = 78; and the area is (1/2)(48)(78) = (24)(78) = 1872.
And it would have been quicker if I'd realized that both right triangles are integer Pythagoreans: AEC is 14-48-50 and EBC is 3-4-5 x 16. But I didn't notice that until the end.
Cheers. 🤠
La vertical por “C” corta a AB en “E”→ Tomando CE como eje, obtenemos el punto D simétrico de A → ∠ADC=2X → CA=50=CD=DB → AB=AE+ED+DB =b+b+50 → 50²-b² = CE²= 80²-(b+50)² → b=14 → AB=(2*14)+50=78 → CE²=50²-b² →CE=48 → Área ∆ABC =78*48/2=1872
Gracias y saludos.
I used another (a little more complicated) way : sine law to get cos x, then cosine law to get the third side, then Heron's formula to get the area.
At a quick glance, Using the sine rule Sin(x)/50 = sin(2x)/80 = sin(C)/AB and the Trig function Sin(2x)= 2*sin(x) * cos(x) Then sin(x)/sin(x)cos(x)= 100/80 then cos(x) = 80/100 and x= 36.87, 2*x = 73.74 and Angle ACB = 180-3*x =69.4. Then AB = (Sin(69.4)/sin(36.87))*50 = 78. area is half base * height = 39 * sin(36.87) * 80 = 1872 area units.
The content of your channel is exactly what I was looking for! Kudos for the good work sir! Please keep uploading more of these plane geometry problems. Very helpful.
Brilliant exercise, looks simple, and is quite simple, but the answer doesnt just jump out at you. You have to make some effort to work your way to the result and that makes it both interesting and challenging. Only humans I think can plan strategies for problem solving, its a gift we should all try to develop, often frustrating but always rewarding. Thanks again
thankyou so much dear professor ❤
You are very welcome
My first thought was to solve this by using trig. Thanks for showing this other method.
I have solved this problem using trigonometry - a bit different approach. Awaiting for such as more.
You can use Sin2x/80 = Sinx50, and then use the identity Sin2x = 2SinxCosx, from there make Cosx the subject of the formula, it's an easier method for me
Once you determine that x = 14, there's no need to calculate to find h, as [7, 24, 25] is a primitive Pythagorean Triple, and we already have a 14 (7×2) and a 50 (25×2), meaning 48 (24×2) is the other side.
Very good problem and easy solution.
50/sinx = 80/sin2x.
General formula, sin2x = 2.sinx.cosx.
Therefore 50/sinx =80/2.sinx.cosx.
50.2.sinx.cosx = 80.sinx.
100 sinx.cosx = 80.sinx.
Cos x = 80.sinx/100sinx.
Cos x = 0.8.
Cos -1, 0.8 = 36.8698 degrees.
Angle ACB =180 - 3x.
ACB = 180 - 110.6094.
ACB =69.3906 degrees.
Area of triangle = 1/2 x 50 x 80 x sin 69.3906.
1872.
Using the trigonometric identities, we can calculate that the sine of the angle ACB is very precisely 117/125😉
(this without ever going through the angle values in degrees, imprecise even when rounded to the 4th decimal)
Extending AB side on the left to point F so that CF= CB=80. (FCB is isoscele)
angle CFA = x by construnction
angle FCA = x because of the sum of external angle
ACF triangle is also isoscele
so FA = CA = 50
I found the area of ACF triangle with Erone's formula (only to use something different of Pythagora...) and got Area = 1200
then found the height of ACF , CE=(1200*2)/50 = 48
then calculated AE and EB with Pythagorean Theorem... but I prefer trigonometric solution
That is how I did it. Area = 1872.
Cool!
Exactly samething can be done by drawing a line CF angled X to the left of A.
Thank you! Cheers! 😀
7:23 A=sqrt(s(s-a)(s-b)(s-c))
s=(a+b+c)/2=(50+78+80)/2=104
A=sqrt(104×54×26×24)=
sqrt(2^8×3^4×13^2)=
16×9×13=1872
Sin rule for a single angle combined with the Sin rule for the sum of two angles does the job.
80 /50 = sin(2 x°)/sin(x°) = 2 cos(x°)
cos ( x°) = 4/5 , sin (x°) = 3/5
sin (3x°) = 3 sin(x°) - 4 sin^3 (x°)
= 3 x 3/5 - 4 (27)/125
= (225 - 108)/225 = 117/225
= 13/25
Hereby area of triangle
= 80 x 50 x 13 /25 x (1/2) sqr unit
= 1040 sqr unit
Draw a Perpendicular h from C to AB. Sin 2x = h/50. Sin x = h/80. h = 80*Sin x. SIn 2x = 80*Sin x /50. 2(Sin x)*(Cos x) = 1.6*Sin x. Cos x = 0.8 = EB/80. EB = 64.
Cos 2x = 2Cossqx - 1 = .28 = AE/50. AE = 14. AB = 14+64 = 78. x 36.87. h = 80*Sin36.87 = 48. Area = 0.5*48*78 = 1872
Thank you! Cheers! 😀
Καλημέρα σας. Μπορούμε να πάρουμε το επεκτεταμένο (Γενικευμένο) Πυθαγόρειο θεώρημα για την αμβλεία γωνία BDC του ισοσκελούς τριγώνου BDC και να υπολογίσουμε το x. Μετά με Πυθαγόρειο θεώρημα στο CDΕ, υπολογίζουμε το h. ...
50/80=sinx/sin2x, sinx=3/5
Area=½.80.50sin3x
=2000(3sinx-4sin³x)
=2000(9/5-4×27/125)
1872
Thanks for video.Good luck sir!!!!!!!!!
Probably better use a diff letter for the length , as x already used as the angle
amazing problem 😊like always 😊
Thank you so much 😀
For grade 11. Using trigonometric functions could be saving some steps.
Sin2x=2sinx.cosx
50sin2x = 80sinx
Cosx=4/5
Sinx=3/5
Cos2x = 7/25
Hello. CEB triangle is 3-4-5 triangle. Sixteen times it. So answer is shorter by this way.😊
After seeing the right triangles formed, the problem became easier
Yay! I solved it.
I agree with th solutions shown, but I disagree with the fact that AB segment is smaller than the CB segment (particularly, 74 versus 80) as seen on the drawing. Whoever came with this problem should have used real values for angles to calculate the sides, and use the real proportions between the two sides given ( AC and BC).
The values have to make sense too.At least for me...
Clear and well-presented.
thank you!
You're welcome!
When i was typing my rule of sines method.,i tried to show you the calculation of Area by three differnt ways. I pressed something and everything was lost. Luckily MrPwmiles solved it by one of my three ways. I'll discribe the other two ways. First Trigonometrical an second geometrical.Rule of Sines gave
sin (2x)=80*sinx/50 so 2*sinx*cosx=80*sinx/50 so cosx=80/100=4/5 and sinx=3/5. Formula of Area of Triangle is **1/2*a*b*sin(
Dal teorema dei seni risulta cosx=4/5...A=(50cos2x+80cosx)*80sinx/2=24*78
Nice problem.
Good probleam.i like it😎
Area=(78×48)/2=1872sq unit.
Thanks sir
So nice of you
Thank you! Cheers! 😀
Good morning sir
About sins rule 👌 😌
sin x=50 , sin 2x=80
Sin (x/50)=2sinx ×cosx /80
Cosx=80/100
x=54
Thank you! Cheers! 😀
not sin(x)=8/10, but cos(x)=4/5 and x= d36.87
Oops! sin x / 50 = 2 sin x cos x / 80 ⇒ cos x = 80/100 (not sin x = 80/100) ⇒ x = 36.87° (the other angle is 53.13°)
@@ybodoN yeah but approximately sinx=4/5 x=54 sorry 🤣
@@alinayfeh4961 the problem is not the wrong rounding but the wrong angle (because a "cos" mistakenly became a "sin")
Strange that the actual triangle looks SO different from the diagram. Sure "not to scale" etc., but usually they look approximately as they should...
h^2=50^2-p^2=80^2-(p+50)^2
p=14
Area of triangle=(2p+50)*48/2=(28+50)*48/2=1872 square units
Thank you! Cheers! 😀
Because the sum of theinterior angles of triangle CAD equals 180⁰ and angles CDA and CDB make a straight angle 180⁰...so subtract angle CDA from both interior triangle and straight angles and wallah! The sum of opposite interior angles. 🙂
Thank you! Cheers! 😀
Sin(2x)=2sin(x)cos(x)...thats the starting point. The area is (14)(48)/2+(64)(48)/2.
So h/50 = 2(h/80)(EB/80)
The h cancels out, allowing for a quick solve for EB. The rest of the triangle then just comes together using the pythagorean theorem.
Using Sin and Cos is more straight forward.
∆ABC → AB = AS + BS; AC = 50; BC = 80; sin(ASC) = 1; CS = h; CAB = 2ϑ; ABC = ϑ; area ∆ ABC = ?
sin(2ϑ) = 2sin(ϑ)cos(ϑ) = h/50 → h = 100sin(ϑ)cos(ϑ); sin(ϑ) = h/80 → h = 80sin(ϑ) = 100sin(ϑ)cos(ϑ) →
cos(ϑ) = 4/5 → sin(ϑ) = 3/5 → sin(2ϑ) = 24/25 → cos(2ϑ) = 7/25 → ∆ BCS = pyth. triple 16(3 - 4 - 5)
∆ ASC = pyth. triple 2(7 - 24 - 25) → h = 48 → BS = 64 → AS = 14 → AB = 78 → area ∆ ABC = 39(48) = 13(12)^2
Its easy with sin rule❤
Why I become 2000? I find area of square and divide on 2
Sir iss question ko maine vector ka concept lagake solve kiya
Bravo!
Thank you! Cheers! 😀
Kaise mujhe jara hints dijiye...
Bro woh wla method Jahangir angle bane wha cosx Lena Hai use perpendicular mein sinx top base aur height ko sinx and cost substitute karke solve kiya
Resolution of vector se
6x=180 x=30 ... error
Out box? First I would do it in a traditional way. Draw CD, where D on AB and angle ADC 2x, thus the triangle DBC is an isosceles triangle with sides 50-80-50, then clearly the height of this triangle is 30, and sin x is 3/5, and the angle outside ACB is 3x, the height of the great triangle is 50xsin 3x=50(3sin x-4sin^3 x)=50(9/5-108/125)=90-43.2=46.8, therefore the answer is 80x46.8/2=1872.🙂
Thank you! Cheers! 😀
@@PreMath My solution is still imcomplete, now it is.😃
@@misterenter-iz7rz
Great!
Good morning sir
Same to you