I solved it by dropping an altitude from A to BC at D. Then two equations. Sin(2x) = y/3 which becomes 2 sin x cos x = y/3 AND sin x = y/5. Substituting gives CD = 25/6. Pythagorean theorem to solve for length of altitude. 5 sqrt (11)/6. Now calculate DB using Pythagorean theorem again. It is 7/6. So, now you have perpendicular line segments so you can use A = 1/2 LW. A = 20 sqrt (11)/9 which equals your approximate 7.37.
Alternative way: Draw a line from B to AC that divides angle ABC (2x degrees) in two equal angles of x degrees. Call the intersection of this line with AC point D. You now have an isosceles triangle BCD with base CB and two angles of x degrees (angle DCB and angle CBD). That means sides CD and BD are of equal length so CD=BD. The angle CDB (top angle) equals 180-2x (both base angles are x degrees). The side CD of triangle BCD is part of the side AC of triangle ABC so both angles at point D (angle CDB and angle ADB) must add up to 180 degrees. With angle CDB=180-2x that means angle ADB=180-*180-2x)=2x. And so triangles ABD and ABC are similar triangles, both have an angle of x degrees and an angle of 2x degrees and share the top angle (angle CAB=angle DAB). You can now state that AC/AB=AB/AD which gives 5/3=3/AD and so AD=9/5. That means CD=AC-AD=5-9/5=16/5 and since CD=BD that means BD=16/5. You can also state that BC/BD=AC/AB which gives BC/(16/5)=5/3 and so BC=16/3. Now all three sides are known, AC=5 (b), AB=3 (c) and BC=16/3 (a). The perimeter of ABC then is a+b+c=16/3+5+3=40/3 and so half the perimeter s=p/2=20/3. The area of a triangle can be calculated using A=√(s*(s-a)*(s-b)*(s-c)) and we get A=√(20/3*4/3*5/3*11/3)=√(4400/81)=(20/9)*√11, approximately 7.37.
Another way to get the length of CB is to use the cos rule applied to angle C. We know Cos(C) = 5/6 (watch first part of video). This give a quadratic equation for the length BC (call it x): (3x-16)(x-3) = 0. x can't be 3 as that has the wrong angle ab B so x = 16/3.
Awesome way to solve it, thanks a lot! 😊. Law of sines is very helpfull. Did it the traditional way, BC = 16/3; h = 5√11/6; area triangle CBA = 20√11/9 square units.
Area can be solved even without calculating the angles (without calculator). Knowing cos(x)=5/6 from your calculation we can calculate sin(x)=sqrt(1-cos2(x))= sqrt(11)/6 Using cos(2x)=cos2(x) - sin2(x) we can calculate Area=1/2(base x hight) base=5cos(x) + 3cos(2x) = 192/36 hight = 5sin(x) = 5sqrt(11)/6 Therefore Area=20sqrt(11)/6 =7,370277312....
I like those trigonometric problems (law of sines, law of cosines, area of triangle formula) please more of that or more of the sin-cos conversions and similar stuff (Sin (2x) = 2 sin x cos x … etc). Very interesting! (They didn‘t teach us that in school…)
Thankuuu thankuu so so so much Sir for this amazing question ... My eyes were really waiting for Trigonometry questions from you ... 🙏🙏❤️❤️❤️❤️❤️❤️❤️❤️ Thanks for your help ....
Here's a slightly different approach, not requiring calculation of any angles, or even the law of sines: Drop a vertical from A to the base. Call it Q. This divides the base into two segments; call them R (left) and S (right). Now Q = 5 sin x = 3 sin 2x. Then by the double-angle formula, 5 sin x = 3 (2 sin x cos x) = 6 sin x cos x. Divide both sides by 6 sin x and we have (5 sin x / 6 sin x) = (6 sin x cos x) / (6 sin x); simplifying, it becomes cos x = 5/6. We now calculate sin x using the identity sin^2 x + cos^2 x = 1: sin x = sqrt(1 - cos^2 x) = sqrt(1 - 25/36) = sqrt(36/36 - 35/36) = sqrt(11/36) = sqrt(11) / 6. Now Q = 5 sin x = (5)(sqrt(11) / 6) = 5 sqrt(11) / 6; R = 5 cos x = (5)(5/6) = 25/6; and S = 3 cos(2x); using another double-angle formula, S = 3 cos(2x) = (3)(2 cos^2(x) - 1) = 3 cos^2(x) - (3)[((2)(25/36) -1)] = (3)[(50/36) - 1)] = (3)[(50 - 36)/36) - (36/36)] = (3)(14/36) = 42/36 = 7/6. So Q (the altitude) is 5 sqrt(11)/6; and R + S (the base) is 25/6 + 7/6 = 32/6; so the area is (1/2)(Q)(R+S), or (1/2)((5 sqrt(11)/6)(32/6) = (1/2)(160 sqrt(11) / 36) = 160 sqrt(11) / 72 = 20 sqrt(11) / 9; Which calculates out to approximately 7.37 square units, the same as your answer. No calculating angles--in fact, no calculators, except to compare answers at the end--no law of sines; and not even the Pythagorean theorem, except as it underlies some of the identities. This is actually very similar to the last problem I did, just before lunch; but the triangle is reversed and the sides are all divided by six. Thank you. ladies and gentlemen; I'll be here all week.
From point A drop down 2 lines to BC. 1) AD so that angle CAD=x. 2) AE so that AEB=90 degrees. Angle CAD=x. So angle ADB=2x. AD=AB=3. With PYTH you find DB=EB=7/6. And Pyth gives you height AE=2.76. Area = AE times BC/2= 7.36
Другое решение с применением тригонометрии: Пусть Sin2x/sinx=5/3, => cosx=5/6, Sinx=V11/6. CB= 3sin3x/sinx=3(Sin2x*cosx+cos2x*sinx)/sinx=3(2(Cosx)^2+cos2x)=3(4(cosx)^2-1)=3*(4*25/36-1)=16/3, 3/sinx=2R => 4R=36/V11. A=a*b*c/4R=5*3*(16/3)*V11/36=20V11/9=7,37. Ответ: 20*V11/9; 7,37 приближенно. Мне нравятся оба способа. В нашей школе на экзаменах не применяют приближенные вычисления.
Sin(180-3x) = sin 3x and sin of 3x = 3sinx-4sin cubed x = 20 by sq root of 11/9. To find x we get 5/sin2x = 3/sin x .That gives cos x =5\6 So sin x=sq root 11 \6 using a rt angled triangle. You don't need to use a calculator.
you don't have to resort to calculator that early ... use law of sines and double angle formula (as you did) to get cos(x), can then get sin(x) as sqrt(1-cos(x)^2), then use area formula for (1/2)ab*sin(180-3x) == (1/2)ab*sin( 180-x - 2x ) -- use trig identities for sine difference of angles, and identities for sin or cos (pi-x) and double angle formula to get closed form (1/2)*a*b*( sin(x)*(cos(x)^2 - sin(x)^2) + cos(x)*(2*sin(x)*cos(x)) ) whole lot more trig going on this way :D
Great method which I used to get the answer for the previous video. However I used my calculator which uses Newton's method to solve equations to find the angle value. Now at least I know the theory and use of double angle formula.
5/3 = sin(2x)/sin x = 2 sin(x) cos(x) / sin(x) = 2 cos x => cos x = 5/6, so sin^2 x = 1 - 25/36 = 11/36 A = 180 - (x + 2x), so sin A = sin 3x = 3 sin x - 4 sin^3 x = (4 cos^2 x - 1) sin x = (4 x 25/36 - 1 = 16/9) (√11/6) = 8 √11 / 27 Area = bc/2 sin A = bc/2 sin (180-A) = bc/2 sin(B+C) = bc/2 sin(2x+x) = bc/2 sin(3x) = 5x3/2 x 8 √11 / 27 = 20 √11 / 9 [ approximately 7.370277311900889... ]
I push dislike button when i know you using Scientific-Calculator* 🚫 Sorry. You should count A corner = 180-3x ! sin A = sin (180-3x) = sin x • [3 - 4.sin"x]* = (8/27).√11 Area = (1/2).5.3.(8/27).√11 = (20/9).√11 = 7,37 cm" !! 😉
Thank you sir . Thanks a lot !
did you pin this comment yourself??????
I solved it by dropping an altitude from A to BC at D. Then two equations. Sin(2x) = y/3 which becomes 2 sin x cos x = y/3 AND
sin x = y/5. Substituting gives CD = 25/6. Pythagorean theorem to solve for length of altitude. 5 sqrt (11)/6. Now calculate DB using Pythagorean theorem again. It is 7/6. So, now you have perpendicular line segments so you can use A = 1/2 LW.
A = 20 sqrt (11)/9 which equals your approximate 7.37.
Alternative way:
Draw a line from B to AC that divides angle ABC (2x degrees) in two equal angles of x degrees. Call the intersection of this line with AC point D.
You now have an isosceles triangle BCD with base CB and two angles of x degrees (angle DCB and angle CBD). That means sides CD and BD are of equal length so CD=BD. The angle CDB (top angle) equals 180-2x (both base angles are x degrees).
The side CD of triangle BCD is part of the side AC of triangle ABC so both angles at point D (angle CDB and angle ADB) must add up to 180 degrees. With angle CDB=180-2x that means angle ADB=180-*180-2x)=2x. And so triangles ABD and ABC are similar triangles, both have an angle of x degrees and an angle of 2x degrees and share the top angle (angle CAB=angle DAB).
You can now state that AC/AB=AB/AD which gives 5/3=3/AD and so AD=9/5. That means CD=AC-AD=5-9/5=16/5 and since CD=BD that means BD=16/5.
You can also state that BC/BD=AC/AB which gives BC/(16/5)=5/3 and so BC=16/3.
Now all three sides are known, AC=5 (b), AB=3 (c) and BC=16/3 (a). The perimeter of ABC then is a+b+c=16/3+5+3=40/3 and so half the perimeter s=p/2=20/3. The area of a triangle can be calculated using A=√(s*(s-a)*(s-b)*(s-c)) and we get A=√(20/3*4/3*5/3*11/3)=√(4400/81)=(20/9)*√11, approximately 7.37.
Thanks for sharing
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Another way to get the length of CB is to use the cos rule applied to angle C. We know Cos(C) = 5/6 (watch first part of video). This give a quadratic equation for the length BC (call it x): (3x-16)(x-3) = 0. x can't be 3 as that has the wrong angle ab B so x = 16/3.
Awesome way to solve it, thanks a lot! 😊. Law of sines is very helpfull.
Did it the traditional way,
BC = 16/3; h = 5√11/6; area triangle CBA = 20√11/9 square units.
Area can be solved even without calculating the angles (without calculator). Knowing cos(x)=5/6 from your calculation we can calculate sin(x)=sqrt(1-cos2(x))= sqrt(11)/6
Using cos(2x)=cos2(x) - sin2(x) we can calculate
Area=1/2(base x hight)
base=5cos(x) + 3cos(2x) = 192/36
hight = 5sin(x) = 5sqrt(11)/6
Therefore Area=20sqrt(11)/6 =7,370277312....
I like those trigonometric problems (law of sines, law of cosines, area of triangle formula) please more of that or more of the sin-cos conversions and similar stuff (Sin (2x) = 2 sin x cos x … etc). Very interesting! (They didn‘t teach us that in school…)
Thankuuu thankuu so so so much Sir for this amazing question ... My eyes were really waiting for Trigonometry questions from you ... 🙏🙏❤️❤️❤️❤️❤️❤️❤️❤️ Thanks for your help ....
شكرا لك على المجهودات من المغرب.
Very important sharing so thank you
Here's a slightly different approach, not requiring calculation of any angles, or even the law of sines:
Drop a vertical from A to the base. Call it Q. This divides the base into two segments; call them R (left) and S (right).
Now Q = 5 sin x = 3 sin 2x. Then by the double-angle formula,
5 sin x = 3 (2 sin x cos x) = 6 sin x cos x.
Divide both sides by 6 sin x and we have
(5 sin x / 6 sin x) = (6 sin x cos x) / (6 sin x); simplifying, it becomes cos x = 5/6.
We now calculate sin x using the identity sin^2 x + cos^2 x = 1:
sin x = sqrt(1 - cos^2 x)
= sqrt(1 - 25/36)
= sqrt(36/36 - 35/36)
= sqrt(11/36)
= sqrt(11) / 6.
Now Q = 5 sin x = (5)(sqrt(11) / 6) = 5 sqrt(11) / 6;
R = 5 cos x = (5)(5/6) = 25/6; and
S = 3 cos(2x); using another double-angle formula,
S = 3 cos(2x) = (3)(2 cos^2(x) - 1)
= 3 cos^2(x) - (3)[((2)(25/36) -1)]
= (3)[(50/36) - 1)]
= (3)[(50 - 36)/36) - (36/36)]
= (3)(14/36)
= 42/36
= 7/6.
So Q (the altitude) is 5 sqrt(11)/6; and R + S (the base) is 25/6 + 7/6 = 32/6;
so the area is (1/2)(Q)(R+S), or
(1/2)((5 sqrt(11)/6)(32/6)
= (1/2)(160 sqrt(11) / 36)
= 160 sqrt(11) / 72
= 20 sqrt(11) / 9;
Which calculates out to approximately 7.37 square units, the same as your answer. No calculating angles--in fact, no calculators, except to compare answers at the end--no law of sines; and not even the Pythagorean theorem, except as it underlies some of the identities.
This is actually very similar to the last problem I did, just before lunch; but the triangle is reversed and the sides are all divided by six.
Thank you. ladies and gentlemen; I'll be here all week.
Awesome, very elegant! 🙂
Thank you very much. I have liked it.
From point A drop down 2 lines to BC. 1) AD so that angle CAD=x. 2) AE so that AEB=90 degrees. Angle CAD=x. So angle ADB=2x. AD=AB=3. With PYTH you find DB=EB=7/6. And Pyth gives you height AE=2.76. Area = AE times BC/2= 7.36
Same as the previous problem.
Area ½.5.3(.√11/6)16/9
=20√11/9=7.37
Awesome as always.
very well Sir
thank you so much Sir ❤
Другое решение с применением тригонометрии: Пусть Sin2x/sinx=5/3, => cosx=5/6, Sinx=V11/6. CB= 3sin3x/sinx=3(Sin2x*cosx+cos2x*sinx)/sinx=3(2(Cosx)^2+cos2x)=3(4(cosx)^2-1)=3*(4*25/36-1)=16/3, 3/sinx=2R => 4R=36/V11. A=a*b*c/4R=5*3*(16/3)*V11/36=20V11/9=7,37. Ответ: 20*V11/9; 7,37 приближенно. Мне нравятся оба способа. В нашей школе на экзаменах не применяют приближенные вычисления.
degrees and radsians?
Nice haha I really like your problems as they are interesting geometry! Good job my friend! Haha
To find the rib BC you should to use the cosinus theorem! 😀😉
Sin(180-3x) = sin 3x and sin of 3x = 3sinx-4sin cubed x = 20 by sq root of 11/9. To find x we get 5/sin2x = 3/sin x .That gives cos x =5\6 So sin x=sq root 11 \6 using a rt angled triangle. You don't need to use a calculator.
you don't have to resort to calculator that early ... use law of sines and double angle formula (as you did) to get cos(x), can then get sin(x) as sqrt(1-cos(x)^2), then use area formula for (1/2)ab*sin(180-3x) == (1/2)ab*sin( 180-x - 2x ) -- use trig identities for sine difference of angles, and identities for sin or cos (pi-x) and double angle formula to get closed form
(1/2)*a*b*( sin(x)*(cos(x)^2 - sin(x)^2) + cos(x)*(2*sin(x)*cos(x)) )
whole lot more trig going on this way :D
sin(180-3x) = sinx(3-4(sin^2)x) then just use the value of sin x next and calculate
the EXACT value of A. No need for calculators
Awesome explanation sir
But i want to share a question with you so how can i ?
That’s great ! But I just want to know how we can fine angle x ? Mostly while we exam no one used calculator allowed 🙂
3/sin x = 5/ sin 2x; 3/ sin x = 5/ 2 sin x cos x; cos x = 5/6, sin x = (√11)/6; S=½ * 5 * 3 * sin (п-3x) = (15/2)* sin 3x.
That’s a great idea 💡 but it make me still not clearly sir
Great method which I used to get the answer for the previous video. However I used my calculator which uses Newton's method to solve equations to find the angle value. Now at least I know the theory and use of double angle formula.
thanks so much!!!!!!!!!!!!!
5 Sin x = 3 Sin 2x
5 Sin x = 3 . 2 Sin x Cos x
5 = 6 Cos x
#Trigonometry
#sine
Same as the 30::18 problem
7½>area
#LawofSines
🙏🏿🙏🏿🙏🏿sir
Angelo in B = 2x, = 67.52
5/3 = sin(2x)/sin x = 2 sin(x) cos(x) / sin(x) = 2 cos x => cos x = 5/6, so sin^2 x = 1 - 25/36 = 11/36
A = 180 - (x + 2x), so sin A = sin 3x = 3 sin x - 4 sin^3 x = (4 cos^2 x - 1) sin x = (4 x 25/36 - 1 = 16/9) (√11/6) = 8 √11 / 27
Area = bc/2 sin A = bc/2 sin (180-A) = bc/2 sin(B+C) = bc/2 sin(2x+x) = bc/2 sin(3x) = 5x3/2 x 8 √11 / 27 = 20 √11 / 9 [ approximately 7.370277311900889... ]
First view
#cosine
Angle A = 180° - 100.68°
Very critical solve cannot understand
A few of the questions have errors and therefore can’t be solved by any method. Please check before posting on this platform.
sin (2x) = 2 × sin (x) × cos (x)
i found area = (20\9) *square11
#DoubleAngleFormula
x = acos (5÷6)
cos (x) = 5 ÷ 6
cos (x)
x = 33.56°
x variable
sin (79.32°)
sin (x) ÷ 3
Sine
sin (x)
I push dislike button when i know you using Scientific-Calculator* 🚫 Sorry.
You should count A corner = 180-3x !
sin A = sin (180-3x) = sin x • [3 - 4.sin"x]* = (8/27).√11
Area = (1/2).5.3.(8/27).√11 = (20/9).√11 = 7,37 cm" !! 😉
What is this crap with the approximate calculations??! if you know that cos x = 5/6, then you can get precise answer - S = 20*SQRT(11)/9
O
very well Sir
thank you so much Sir ❤