Two Methods: With & Without Trigonometry | Find the Length of AB & AC in this Triangle

[Jack_Callcott_AU]

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.

[DxRzYT]

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)

[patrickjacquiot]

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.

[wtspman]

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.

[arthurschwieger]

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.

[geoellinas]

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.

[guneesh]

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

[a.n.timoon-sb]

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.

[michaelstubbs]

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.

[difana]

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

[fongalex]

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

[jakkima]

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 ;)

[malcolmmcgrath]

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.

[williamwingo]

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.

[suheochanel]

I love studying math. thank you for sharing. You have a lot of knowledge about math. keep connected.

[eduardoteixeira]

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

[navink]

Thanks PreMath I’m enjoying your videos. After 15 years I’m learning math again 🙏

[Volti]

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

[ssa]

По теореме о сумме углов треугольника /_А=180°-(45°+30°)=105°.
По теореме синусов:

АВ=8

[yegnanarayana]

Instead of using phythogorous theorem we canuseDB=Square root of 3times x square root of 3x+X=1.732x=8 and then solve for