On peut utiliser aussi le produit scalaire : V(AB) . V(AC) = AB x AC x cos (ABC) et V(AB) . V(AC) = 1/2 (AB^2 + AC^2 - BC^2).

In Method 2, @ 3:30, we can't assume that θ is obtuse (ie., greater than 90°). Since θ is flanked by sides of a=2 & b=3, we can check the square of the length of the third side c²=(√19)²=19 against a²+b²=2²+3²=4+9=13. Since c²>a²+b², we have θ>90°. This means that when we drop an altitude from point B to line containing segment AC, it intercepts line AC external to the triangle.

@jimlocke9320

2 ай бұрын

Furthermore, with Θ known to be >90°, if we drop a perpendicular from C to the line containing AB, it will intercept external to the triangle, call the intersection point E. We make our educated guess that ΔACE is a special 30°-60°-90° right triangle and try AE = 1.5, half of hypotenuse AC, Then, CE = 1.5√3 and BE = 3.5. (BE)² + (CE)² = (3.5)² + (1.5√3)² = 12.25 + (6.75) = 19, which is side BC squared. So

Method #1, law of cosines is, obviously, the straightforward way to solve but method #2 works well for those who choose not to memorize law of cosines. Method #2 also holds a shortcut. A good educated guess is that ΔABD is a special triangle and we try 30°-60°-90°, guessing that x = 1 and y = √3. We test the Pythagorean theorem and find that right ΔBCD with sides 4 and √3 does have a hypotenuse of √19. From there, it is straightforward to find that Θ = 120°.

Solution with PURE Geometry. Obviously BC² > AB²+AC² => θ>90° . Apply generalization of the Pythagorean theorem in triangle ABC. BC²=AB²+AC²+2•AC•AD => 19=4+9+ 2•3AD => AD=1 . In orthogonal triangle AD=AB/2 =>

🇧🇷 Bela explanação de ambos os métodos. Obrigado.

New York Trigonometry Regents-June,1961.

❤❤❤❤ thanks 💯💯 💯💯

Boh,con le formule di Briggs risulta cosθ/2=1/2...θ=120...ah,il primo disegno non era corretto

@pieterjlansbergen6988

2 ай бұрын

Pork.. addirittura le formule di Briggs.. Ma perché complicarsi così la vita? Con il teorema del coseno la soluzione è banale. 😊

@giuseppemalaguti435

2 ай бұрын

​​@@pieterjlansbergen6988la formula di Briggs ti dà subito il risultato...non capisco qual è il ptoblema... ovviamente come la formula del coseno,infatti tutte le formule trigonometriche sui triangoli sono una derivante dall'altra..