May I suggest that in the second part " Without Trigonometry" it's a little bit easier to say (8-x)/x =sqrt(3) by "special triangles " and solve for x thus avoiding using the theorem of Pythagoras and then the quadratic equation formula. Thanks for the video.
Great video! Before watching the video, I used the Sine Rule to calculate this. Since 2 angles are given to us in a triangle, we can calculate the missing angle as angles in a triangle add up to 180 (which gives you 105°) Then just use the sine rule to find both AC and AB. 8/Sin(105°) = AB/Sin(45°) rearrange: Sin(45°) × 8/Sin(105°) = AB AB = 5.9 (1.d.p) You can use the cosine rule here, but I just used the sine rule again as it's simpler. 8/Sin(105°) = AC/Sin(30°) Rearrange [Sin(30°) is ½) ½ × Sin(105°) = AC AC = 4.1 (1.d.p)
You didn’t need to use Pythagoras for working out x. Side BD is sqrt(3)*x from 30-60-90 triangle rule. BC = x + sqrt(3)*x = 8. Rearraging and rationalizing the fraction gives (8(sqrt(3)-1))/2 = 4(sqrt(3)-1). No Pythagoras or quadratic formula needed.
I am a Greek (My great-grandfather Pythagoras :) and our teachers told us to follow the shortest path. So based on this recommendation, I have the very short solution for the non-geometric method without we write the straight line segment DB as 8-x! Using Pythagorean thm in the triangle ADB we find DB=sqrt(3)*x, so CD+DB=8 => x+sqrt(3)*x=8 => [(1+sqrt(3)]*x=8 => x=8/[1+sqrt(3)]. So we avoiding the use of the quadratic equation.
Here is a third way to do this. ADB is a 30/60/90 so DB is X * √3 based on the relationship between the sides of a 30/60/90. That means 8-X = X * √3. This gives us X = 8/((√3 + 1) = 2.93. Then for side AB, we have 2 * 2.93 = 5.86. That also gives us side AC of 2.93 * √2 = 4.14. Thank you for providing these math problems. I like the mental workout.
Exactly. I made the same comment. Set x times the square root of 3 equal to 8 minus x. Then simply solve for x (2.9282), and thereby skip all the unnecessary algebra.
Thank you for all these exercises you have created, they certainly exercise my creaky brain. I am looking forward to a time that I can actually solve one prior to following your instructions. With regard to this exercise, in the non trigonometry solution at 14:53 x=-4+4√3 becomes x=4√3-4 . Could you please explain how that sign shift is determined.
There is another special triangle namely 30-60-90. You can use both triangles. x+x times square root3=8. Then x = 8/(1+root3) is approximately 2.93. Then AB is 2 times 2.93 is 5.86 and AC is square root 2 times 2.93 is 4.14.
I figured that 8 is about 2.73X and x equals to about 2.92 because of something in 8:20 minute. if you notice the left side of the triangle is "a" times square root of 3 but a is actually "x". 8 = x + x*sqrt(3) = 2.7x you get that x is 2.92 and from that you get the other numbers
Here's a sort of "combination" solution: In the 30-60-90 triangle (ABD), if the side opposite the 30 degree angle is 1, then the hypotenuse is 2 and the adjacent side is sqrt(3). This follows from the fundamental identity sin 30 degrees = 1/2, and the Pythagorean theorem. So if opposite side is x and the adjacent side is 8 - x, we have x sqrt(3) = 8 - x; add x to both sides: x sqrt(3) + x = 8; collect terms: x (sqrt(3) + 1) = 8; divide both sides by (sqrt(3) +1): x = 8 / (sqrt(3) +1); remembering that the product of the sum and the difference of the same two numbers is the difference of their squares, we multiply top and bottom by (sqrt(3) -1) and get x = 8 (sqrt(3) -1) / ((sqrt(3) + 1) (sqrt(3) -1)) = 8 (sqrt(3) -1) / (3 - 1) = 8 (sqrt(3) -1) / 2 = 4 (sqrt(3) -1) = 4 sqrt(3) - 4. So side AB = 2x = 2 (4 sqrt(3) - 4) = 8 sqrt(3) - 8; and side AC = x sqrt(2) = sqrt(2) (4 sqrt(3) -4) = 4 sqrt(2) (sqrt(3) -1). No calculator, no trig tables (probably hard to find nowadays anyhow), no law of sines, and no quadratic formula. Thank you, ladies and gentlemen; I'll be here all week.
Большое спасибо за подробное объяснение. В нашей школе применяют 1 способ в текущей работе, а второй на экзамене, там нельзя применять калькулятор. На экзамене ответы были бы: 8*(V3-1); 4*(V6-V2).
I do not know if already exist in other commentary the following sugestion In the first solution using trigonometry if we observe that 105 is the same as 60+45 it is possible to calculate the triangle side values without using decimal aproximations. Thank you
По теореме о сумме углов треугольника /_А=180°-(45°+30°)=105°. По теореме синусов: АС=8sin30°/sin105°=....=4/cos15°; АВ=8 sin45°/sin105°=...=4\|2/cos15°.
AB = 2x is the same as saying sin(30°)=1/2 so it is exactly the same method as using trigonometry after all. I would notice that 60° on the top will create an equilateral triangle with a new point E that is on AB at x distance from A. Then triangle DBE is isosceles with equal sides equal to x, making AB=2x. Though 30-60-90 comes from the same idea, so I guess it's still the same. But, they don't teach 30-60-90 theorem in Polish schools so I had to do without it ;)
That's exactly what I thought. I think that your method of proving AB=2x via constructed equilateral and isosceles triangles is the proper way to do it without trigonometry.
Draw AD perpendicular to CB- Let CD = x DB = 8- x In Triangle ADC:- tan 45 = AD/x AD= x In Triangle ADB:- tan 30 = AD/ 8-x = x/8-x So, x/8-x = 1/sqrt(3) x* sqrt(3) = 8-x x = 8/ sqrt(3) +1 x = 2.928 So, CD = 2.928 and DB = 5.072 Sin 45 = 2.928/ AC AC = 4.14 Sin 30 = 2.928/AB AB = 5.856 ~ 5.86
I returned to this problem with the accumulated knowledge from your problems to this point and decided to construct a vertical AD and label it (X). You then have a 45,45,90 triangle on the left and a 30,60,90 triangle on the right with a common side (X). When you mark the side ratios on the left you have X, X, XRoot2 on the left and X , XRoot3 and 2X on the right. You can now find the relationship CD + DB = 8 ; X + XRoot3 =8. When you solve this equation you get X = 4(Root3 - 1). Therefore AB = 8(Root3- 1) and AC = 4(Root6 - Root2). AC = 4.141 ; AB = 5.856.
AB=5.85640.... and AC= 4.14110....... However, very easy question. I needn't think more about it. ( Actually I know the rule a/sinA= b/sinB= c/sinC =2R. Though I am a 10th grade student, I am solving to 11 or 12 grade problems. So, I know the theory. Good question. Thanks.
The Both methods are shows difficult Another simple & easy way to find AC, AB In triangle ADC 45° 45° 90° Let AD = X, CD = X, AC = X√2 In triangle ADB Let AD = X, DB = X√3, AB = 2X In triangle ABC CB. = ( CD+DB ) = 8 ( CD + DB) = X + X√3. = 8 X = 8 ÷ 2.732 = 2.93 AC = X√2. = 2.93 * 1.414 = 4.14 AB = X2 = 2.93*2 = 5.86 Thank you sir.
Why didn’t you set x times the square root of 3 equal to 8 minus x and solve for x? Here x equals 8 divided by 1 plus the square root of 3, or 2.9282. This also equals 4 times the square root of 3 minus 4, but without all the unnecessary algebra.
let AB=x and AC = y 0.5 *y*8*sin45 = 0.5 *x *8*sin30 then x= y square root 2 with Al Kashi you have x2 = y2 +64 -8ysquare2 so 2y2 = y2 +64 -8square 2 y is positive so y = 4square 6 - 4square 2 and x = 8square 3 -8 if i didn t make a mistake
No peeking, no quadratic equations, and no law of sines or cosines: just two special triangles and straight algebra. Drop a perpendicular from A to the base at point D. Both triangles so formed are "special" triangles: ACD is 45-45-90 and ABD is 30-60-90. Let the right segment of the base DB be x and then the left segment CD is 8 - x. The altitude AD is also 8 - x because triangle ACD is isosceles. Then by the properties of the special triangles, length AB is 2(8 - x) [1] and length AC is sqrt(2)(8 - x) [2]. Considering segments AD and DB, we have AD = 8 - x; and also, by the 30-60-90 triangle property: AD = x/sqrt(3); equate the two expressions, 8 - x = x/sqrt(3); multiply both sides by sqrt(3), sqrt(3)(8 - x) = x, multiply out on the left, 8 sqrt(3) - x sqrt(3) = x; 8 sqrt(3) = x + x sqrt(3) = x (1 + sqrt(3)); invert to get all the unknowns on the left, x (1 + sqrt(3)) = 8 sqrt(3); multiply out on the left, x = 8 sqrt(3)/(1 + sqrt(3)); multiply on top and bottom by (sqrt(3) -1), x = 8 sqrt(3)(sqrt(3) - 1)/(3 - 1); multiply out and collect terms, = ((8)(3) - 8 sqrt(3))/2; carry out the division by 2, = ((4)(3) - 4 sqrt(3))/1; = 12 - 4 sqrt(3). Now 8 - x = 8 - 12 + 4 sqrt(3) = 4 sqrt(3) - 4 = 4(sqrt(3) - 1). So using equation [1] above, AB = 2(8 - x) = 8 (sqrt(3) - 1); and from equation [2] above, AC = sqrt(2)(8 - x); = sqrt(2)(4 sqrt(3) - 4); = 4 sqrt(6) - 4 sqrt(2); = (4)(sqrt(6) - sqrt(2)). These are the exact answers. The approximate decimals are AB = 5.8564 and AC = 4.1411, approximately. At this point I ran the video and it looks like we agree. I chose x and 8 - x the opposite way; but then, I'm left-handed. It works out the same either way. Thank you, ladies and gentlemen; I'm here all week.
Draw the triangle with accurate measurements given of angles and the length of bottom line. Angle A should be 105 degrees. If correct,measure the other lines using the same scale you used to measure 8 at the bottom.
First view and first comment
Legend!
Thank you so much Shan dear for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
May I suggest that in the second part " Without Trigonometry" it's a little bit easier to say (8-x)/x =sqrt(3) by "special triangles " and solve for x thus avoiding using the theorem of Pythagoras and then the quadratic equation formula. Thanks for the video.
Great video! Before watching the video, I used the Sine Rule to calculate this. Since 2 angles are given to us in a triangle, we can calculate the missing angle as angles in a triangle add up to 180 (which gives you 105°)
Then just use the sine rule to find both AC and AB.
8/Sin(105°) = AB/Sin(45°)
rearrange:
Sin(45°) × 8/Sin(105°) = AB
AB = 5.9 (1.d.p)
You can use the cosine rule here, but I just used the sine rule again as it's simpler.
8/Sin(105°) = AC/Sin(30°)
Rearrange [Sin(30°) is ½)
½ × Sin(105°) = AC
AC = 4.1 (1.d.p)
I think the best way is using : (sinA)/a=(sinB)/b=(sinC)/c, where A, B, C are angles and a, b, c corresponding opposite sides.
Chuẩn không phải chỉnh
You didn’t need to use Pythagoras for working out x. Side BD is sqrt(3)*x from 30-60-90 triangle rule. BC = x + sqrt(3)*x = 8. Rearraging and rationalizing the fraction gives (8(sqrt(3)-1))/2 = 4(sqrt(3)-1). No Pythagoras or quadratic formula needed.
I also used this method as well.
Можно еще упростить (без Пифагора): sqrt(3)=8-х
I did the same
More well is xV3=8-x. 😀😉
I am a Greek (My great-grandfather Pythagoras :) and our teachers told us to follow the shortest path. So based on this recommendation, I have the very short solution for the non-geometric method without we write the straight line segment DB as 8-x! Using Pythagorean thm in the triangle ADB we find DB=sqrt(3)*x, so CD+DB=8 => x+sqrt(3)*x=8 => [(1+sqrt(3)]*x=8 => x=8/[1+sqrt(3)]. So we avoiding the use of the quadratic equation.
Here is a third way to do this. ADB is a 30/60/90 so DB is X * √3 based on the relationship between the sides of a 30/60/90. That means 8-X = X * √3. This gives us X = 8/((√3 + 1) = 2.93. Then for side AB, we have 2 * 2.93 = 5.86. That also gives us side AC of 2.93 * √2 = 4.14.
Thank you for providing these math problems. I like the mental workout.
Best way to solve, like the most among 3 ways
Exactly. I made the same comment. Set x times the square root of 3 equal to 8 minus x. Then simply solve for x (2.9282), and thereby skip all the unnecessary algebra.
I love studying math. thank you for sharing. You have a lot of knowledge about math. keep connected.
Thanks PreMath I’m enjoying your videos. After 15 years I’m learning math again 🙏
8:08 you can see that: 8-x=x*sqrt(3). from here we can calculate that: x=8/(sqrt(3)+1). the result is the same, but the solution is faster ;)
Nice video! I couldn’t answer this untill you show a possible resolution, you are a great teacher!
Thank you for all these exercises you have created, they certainly exercise my creaky brain. I am looking forward to a time that I can actually solve one prior to following your instructions.
With regard to this exercise, in the non trigonometry solution at 14:53 x=-4+4√3 becomes x=4√3-4 . Could you please explain how that sign shift is determined.
There is another special triangle namely 30-60-90. You can use both triangles. x+x times square root3=8. Then x = 8/(1+root3) is approximately 2.93. Then AB is 2 times 2.93 is 5.86 and AC is square root 2 times 2.93 is 4.14.
That's how I did it. Pretty quick.
A pleasure to read.and follow. Terrific teacher.
I figured that 8 is about 2.73X and x equals to about 2.92 because of something in 8:20 minute.
if you notice the left side of the triangle is "a" times square root of 3 but a is actually "x".
8 = x + x*sqrt(3) = 2.7x
you get that x is 2.92 and from that you get the other numbers
Here's a sort of "combination" solution:
In the 30-60-90 triangle (ABD), if the side opposite the 30 degree angle is 1, then the hypotenuse is 2 and the adjacent side is sqrt(3). This follows from the fundamental identity sin 30 degrees = 1/2, and the Pythagorean theorem.
So if opposite side is x and the adjacent side is 8 - x, we have x sqrt(3) = 8 - x;
add x to both sides: x sqrt(3) + x = 8;
collect terms: x (sqrt(3) + 1) = 8;
divide both sides by (sqrt(3) +1): x = 8 / (sqrt(3) +1);
remembering that the product of the sum and the difference of the same two numbers is the difference of their squares, we multiply top and bottom by (sqrt(3) -1) and get
x = 8 (sqrt(3) -1) / ((sqrt(3) + 1) (sqrt(3) -1))
= 8 (sqrt(3) -1) / (3 - 1)
= 8 (sqrt(3) -1) / 2
= 4 (sqrt(3) -1)
= 4 sqrt(3) - 4.
So side AB = 2x = 2 (4 sqrt(3) - 4) = 8 sqrt(3) - 8;
and side AC = x sqrt(2) = sqrt(2) (4 sqrt(3) -4) = 4 sqrt(2) (sqrt(3) -1).
No calculator, no trig tables (probably hard to find nowadays anyhow), no law of sines, and no quadratic formula.
Thank you, ladies and gentlemen; I'll be here all week.
You have chosen the ideal case for this type of problem:))
Nice work
In general if 2 angles and 1 side are given, use sine law
If 1 angle and 2 sides are given, use cosine law
Sir very good explanation. 🙏🙏
AB=2X AC=X√2
8-X=X√3 AC=8√2/√3+1
X=8/√3+1 AC=8√2(√3-1)/2
AC=4√2(√3-1)
AC=4√6-4√2
Большое спасибо за подробное объяснение. В нашей школе применяют 1 способ в текущей работе, а второй на экзамене, там нельзя применять калькулятор. На экзамене ответы были бы: 8*(V3-1); 4*(V6-V2).
thank you, i never heard this Law befor
thank you so match
Can you show us how to derive the quadratic equation? Thank you!
Yes. Figured it out so quickly by sine law.
But because of rounding, it was a slightly different for me. c= 5,86 cm a = 8,25 cm and b = 4,13 cm
This is great channel perfect for who likes maths like me. You explained very well. Lkd
I do not know if already exist in other commentary the following sugestion
In the first solution using trigonometry if we observe that 105 is the same as 60+45 it is possible to calculate the triangle side values without using decimal aproximations.
Thank you
very well done bro, thanks for sharing
Angle A = 180° - 45° - 30° = 105°
sin 105° = sin 75° = cos 15° = √((1 + cos 30°)/2) = √((1 + √3/2)/2) = √(2 + √3)/2 = (√6 + √2)/4 = 1/(√6 - √2)
AB/sin 45° = AC/sin 30° = 8/sin 105° = 8(√6 - √2)
AB = 8(√6 - √2)*(1/√2) = 8(√3 - 1) = 5.8564
AC = 8(√6 - √2)*(1/2) = 4(√6 - √2) = 4.1411
Я тоже так решаю.
8:09 If a = x, u can see 8 - x = x sqrt(3)
По теореме о сумме углов треугольника /_А=180°-(45°+30°)=105°.
По теореме синусов:
АС=8sin30°/sin105°=....=4/cos15°;
АВ=8 sin45°/sin105°=...=4\|2/cos15°.
AB = 2x is the same as saying sin(30°)=1/2 so it is exactly the same method as using trigonometry after all.
I would notice that 60° on the top will create an equilateral triangle with a new point E that is on AB at x distance from A. Then triangle DBE is isosceles with equal sides equal to x, making AB=2x.
Though 30-60-90 comes from the same idea, so I guess it's still the same. But, they don't teach 30-60-90 theorem in Polish schools so I had to do without it ;)
That's exactly what I thought. I think that your method of proving AB=2x via constructed equilateral and isosceles triangles is the proper way to do it without trigonometry.
3:05 I like that thunderstorm sound, it is cool. I mean, if it is a thunderstorm.
You are correct! Thunderstorms in Arizona, USA!
Love and prayers from the USA! 😃
From your diagram, you can solve it simply by x*SQRT(3)= 8 - x. So that works out to be x = 8/(SQRT(3)+1) = 2.928.
Draw AD perpendicular to CB-
Let CD = x
DB = 8- x
In Triangle ADC:-
tan 45 = AD/x
AD= x
In Triangle ADB:-
tan 30 = AD/ 8-x = x/8-x
So,
x/8-x = 1/sqrt(3)
x* sqrt(3) = 8-x
x = 8/ sqrt(3) +1
x = 2.928
So, CD = 2.928 and DB = 5.072
Sin 45 = 2.928/ AC
AC = 4.14
Sin 30 = 2.928/AB
AB = 5.856
~ 5.86
nice rainy wether at minute 3.27 in window behind speaker in India (or Pakistan) i love it
Impressive solution
So nice of you Subham dear
Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
I returned to this problem with the accumulated knowledge from your problems to this point and decided to construct a vertical AD and label it (X). You then have a 45,45,90 triangle on the left and a 30,60,90 triangle on the right with a common side (X). When you mark the side ratios on the left you have X, X, XRoot2 on the left and X , XRoot3 and 2X on the right. You can now find the relationship CD + DB = 8 ; X + XRoot3 =8. When you solve this equation you get X = 4(Root3 - 1). Therefore AB = 8(Root3- 1) and AC = 4(Root6 - Root2). AC = 4.141 ; AB = 5.856.
After deciding “x” value, if you compare DB, x*root of 3= 8-x, from this we get x directly and easily.
A more rapid solution without solving quadratic equation, using the geometry of the special traingle ADB: x+x*sqrt(3)=8
L'astuce : > AB= Racine (2)× AC
> L'angle (AB, AC) = 105°
>appliquer théorème gle de phytagore au triangle (ABC)
split BC into x+y, such that x+y = 8. y=sqrd(3) * x because tan(60) = sqrt(3).... so on and so forth
AB=5.85640.... and AC= 4.14110....... However, very easy question. I needn't think more about it.
( Actually I know the rule a/sinA= b/sinB= c/sinC =2R. Though I am a 10th grade student, I am solving to 11 or 12 grade problems. So, I know the theory. Good question. Thanks.
Please select difficult problems from previous competitive exams.
it's just √3x+x=8
fantastic
The Both methods are shows difficult
Another simple &
easy way to find AC, AB
In triangle ADC 45° 45° 90°
Let AD = X, CD = X, AC = X√2
In triangle ADB
Let AD = X, DB = X√3, AB = 2X
In triangle ABC
CB. = ( CD+DB ) = 8
( CD + DB) = X + X√3. = 8
X = 8 ÷ 2.732 = 2.93
AC = X√2. = 2.93 * 1.414
= 4.14
AB = X2 = 2.93*2 = 5.86
Thank you sir.
No need for quadratic eq.
AB = x
AH (altitude) = x/2
CH= AH = x/2
HB = sqrt(3)/2*x
(CH+HB) = (x/2 + sqrt(3)/2*x)= 8, . . .
Si reemplazas 8-x por raíz cuadrada de 3 por x, creo es más fácil.
Why didn’t you set x times the square root of 3 equal to 8 minus x and solve for x?
Here x equals 8 divided by 1 plus the square root of 3, or 2.9282. This also equals 4 times the square root of 3 minus 4, but without all the unnecessary algebra.
let AB=x and AC = y
0.5 *y*8*sin45 = 0.5 *x *8*sin30
then x= y square root 2
with Al Kashi you have
x2 = y2 +64 -8ysquare2 so 2y2 = y2 +64 -8square 2
y is positive so y = 4square 6 - 4square 2
and x = 8square 3 -8
if i didn t make a mistake
8-x =sq root 3 multiplied by x. So x = 4 by sq root 3 - 4 etc.
Instead of using phythogorous theorem we canuseDB=Square root of 3times x square root of 3x+X=1.732x=8 and then solve for x```````
Good basic exercise for the brain - not real hard, something like 10 mental push-ups
DB = xsqrt (3) , so CB = x+ xsqrt (3) = 8 than x = 2,93 AB = 5,86 AC = xsqrt (2) = 4,14
Tʜᴀɴᴋs ᴛᴇᴀᴄʜᴇʀ👍
#lawofsines #sine
b = 8 × (sin (30°) ÷ sin (105°)) = 8 × (0.5 ÷ sin (105°)
Use sin law frist to find side b or c.
No peeking, no quadratic equations, and no law of sines or cosines: just two special triangles and straight algebra.
Drop a perpendicular from A to the base at point D. Both triangles so formed are "special" triangles: ACD is 45-45-90 and ABD is 30-60-90.
Let the right segment of the base DB be x and then the left segment CD is 8 - x. The altitude AD is also 8 - x because triangle ACD is isosceles.
Then by the properties of the special triangles, length AB is 2(8 - x) [1] and length AC is sqrt(2)(8 - x) [2].
Considering segments AD and DB, we have
AD = 8 - x; and also, by the 30-60-90 triangle property:
AD = x/sqrt(3); equate the two expressions,
8 - x = x/sqrt(3); multiply both sides by sqrt(3),
sqrt(3)(8 - x) = x, multiply out on the left,
8 sqrt(3) - x sqrt(3) = x;
8 sqrt(3) = x + x sqrt(3) = x (1 + sqrt(3)); invert to get all the unknowns on the left,
x (1 + sqrt(3)) = 8 sqrt(3); multiply out on the left,
x = 8 sqrt(3)/(1 + sqrt(3)); multiply on top and bottom by (sqrt(3) -1),
x = 8 sqrt(3)(sqrt(3) - 1)/(3 - 1); multiply out and collect terms,
= ((8)(3) - 8 sqrt(3))/2; carry out the division by 2,
= ((4)(3) - 4 sqrt(3))/1;
= 12 - 4 sqrt(3).
Now 8 - x = 8 - 12 + 4 sqrt(3)
= 4 sqrt(3) - 4
= 4(sqrt(3) - 1). So using equation [1] above,
AB = 2(8 - x) = 8 (sqrt(3) - 1); and from equation [2] above,
AC = sqrt(2)(8 - x);
= sqrt(2)(4 sqrt(3) - 4);
= 4 sqrt(6) - 4 sqrt(2);
= (4)(sqrt(6) - sqrt(2)).
These are the exact answers. The approximate decimals are AB = 5.8564 and AC = 4.1411, approximately.
At this point I ran the video and it looks like we agree. I chose x and 8 - x the opposite way; but then, I'm left-handed. It works out the same either way.
Thank you, ladies and gentlemen; I'm here all week.
#RightTriangle
Draw the triangle with accurate measurements given of angles and the length of bottom line. Angle A should be 105 degrees. If correct,measure the other lines using the same scale you used to measure 8 at the bottom.
Sine Law. Angle A is 115 degree.
sin 45 not equal that you write ,also sin 105 equal negative you write
#angle
Right triangle ADC
8÷sin(105°) = c ÷ sin (45°)
8-x=x * sqrt(3) and continue
#Trigonometry
Vvnice true thanks
Square root of 192 = 8 × square root of 3
#QuadraticEquation
The 3rd Method，I follow the 2nd Method ，at last I use x/8-x=tag30=√3/3，then x get the answer 。
Use Sin Rule to solve.
#Pythagoras #PythagoreanTheorem
x^(2) + 8x - 32 = 0
Pythagore:(AB×AB)+(AC×AC)
Um…both used trig. Special triangles are trig.
30°; 45°; 105°
sin (30°) = 0.5
sin (45°)
alpha = 105°
Square root of 3
sin (105°)
مجموع درجات المثلث 180 درجة
8-x
Square root of 192
Hello
Nice
Bài toàn này chỉ khoảng 3 phút giải xong.
Is ki length apni garnh oqat say nikalay america dunya ab america ki garnh pa thi
Very lengthy.
Trigo is quicker
8 × square root of 3
alpha + 75° = 180°
x variable
tedious
Nice