I love it when a problem is fully solved both by Geometry and Trigonometry. Thanks for your work!
@SuperPassek
10 ай бұрын
Nice solutions. We can also use vectors. Let b = vector AB and c = vector AC. Then |b| = 6, |c| = 8, |b - c| = 7 |b - c|^2 = |b|^2 + |c|^2 - 2 b • c, so 49 = 36 + 64 - 2 b • c, which means b • c = 51/2 AD = (4b + 3c)/7, in other words, |AD|^2 = (16|b|^2 + 9|c|^2 + 24 b • c) / 49 = 1964 / 49 = 36, so |AD| = 6
@ronaldjorgensen6839
Жыл бұрын
thank you
@markkinnard796
Жыл бұрын
Absolutely Beautiful!
@komedivurdugololdu
5 ай бұрын
Thanks this saved my life
@upalsengupta5878
Жыл бұрын
Excellent👍
@jimlocke9320
Жыл бұрын
Drop perpendicular from A to BC. Label the point of intersection on BC as point E. We now have right triangles ΔABE and ΔACE. Let the common side AE have length h and BE have length x, therefore CE has length 7 - x. From Pythagorean theorem for ΔABE, 6² = x² + h² and for ΔACE 8² = (7 - x) + h². Expand the second equation: 8² = 7² -14x + x² + h² and, from the first equation, substitute 6² for x² + h²: 8² = 7² -14x + 6² 64 = 49 - 14x + 36 14x = 21 x = 1.5 Note that length DE = 3 - x = 1.5 but length BE = x = 1.5, therefore BE and DE have the same length. Therefore, ΔABE and ΔADE are congruent by side angle side (BE and DE have the same length,
@cleiberrocha1449
8 ай бұрын
Wonderful solution!
@syambabuvudara2410
Жыл бұрын
Thank you madam
@marioalb9726
Жыл бұрын
Bisector theorem: x² = 6 . 8 - 3 . 4 x = 6 cm ( Solved √ )
@josephsalinas6725
Ай бұрын
Maravilha !
@じーちゃんねる-v4n
Жыл бұрын
Consider the complex number plane with A as the origin and AD as the real axis. C(8cosθ+i8sinθ) B(6cosθ-i6sinθ) D((48/7)cosθ) CB^2=|2cosθ+i14sinθ|^2=4(cosθ)^2+196(sinθ)^2=196-192(cosθ)^2=49 ∴(cosθ)^2=147/192=49/64 ∴cosθ=7/8 ∴AD=(48/7)(7/8)=6 (JAPANESE) Aを原点、ADを実軸とする複素数平面で考える。 C(8cosθ+i8sinθ) B(6cosθ-i6sinθ) D((48/7)cosθ) CB^2=|2cosθ+i14sinθ|^2=4(cosθ)^2+196(sinθ)^2=196-192(cosθ)^2=49 ∴(cosθ)^2=147/192=49/64 ∴cosθ=7/8 ∴AD=(48/7)(7/8)=6
@Okkk517
Жыл бұрын
How do you arrive to D((48/7)cos(theta))?
@じーちゃんねる-v4n
Жыл бұрын
@@Okkk5171) from angle bisector formula D divides BC into 3:4 so (8x3+6x4)/(3+4)=48/7 2)Since the imaginary axis becomes 0 (3(8c+8si)+4(6c-6si))/7=(48/7)c
@cleiberrocha1449
8 ай бұрын
I prefer the second method.
@MataniMath
Жыл бұрын
I know that you want to show the effect of the angle bisector theorem, then proceed to determine the length of the bisector. But if given the picture looks like that, the length of AD can be determined by cosine rule as follows : In triangle ABC, cos(2t) = (AB^2 + AC^2 - BC^2)/(2(AB)(AC)) = (36+64-49)/96 =17/32. t = theta 2[(cos(t)]^2 - 1 = 17/32 ----> cos(t) = 7/8. In triangle ABD, cos(t) = (36+AD^2 - 9)/[2(6)(AD) = (27 +AD^2)/(12AD). It means, (27+AD^2)/(12AD) = 7/8 ----> 216+8(AD^2) = 84(AD); 2(AD^2) - 21(AD)+54 = 0; [2(AD) - 9][AD - 6] = 0; AD = 4.5 (rejected, not satisfy at triangle ABD) or AD = 6.
@n.662
Жыл бұрын
I found the second method
@himo3485
Жыл бұрын
AD=√(6*8 - 3*4)=√36=6
@valentino1486
Жыл бұрын
It's easier using Stewart's Theorem
@MataniMath
Жыл бұрын
Rightly so; but he wanted to show that the angle bisector theorem is holds, proceeding to determine the length of the bisector
@mathswan1607
Жыл бұрын
Cosine law
@giuseppemalaguti435
Жыл бұрын
Ho calcolato i 3 angoli con le formule di briggs... Ma la loro somma non dà 180.. Boh, are you sure?ok, ho rifatto i calcoli e ho trovato i 3 angoli... Adesso con una qualunque rule sine trovo AD
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