Find the angle X | Geometry Problem | Important Geometry and Algebra skills Explained

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Find the angle X | Geometry Problem | Important Geometry and Algebra skills Explained

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Комментарии
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X =30 Degree You can easily find the answer from the sine area formula.

alicanpolat
Автор

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

اممدنحمظ
Автор

[∆ A B D] = [∆ A C D]
AD being common side of these triangles,
AD.AB sin( BAD) /2
= AD.AC sin( CAD) /2
sin ( CAD) = (AB / AC) sin( BAD)
= 1/2 = sin ( 30° )
angle CAD = x = 30°
sin ( CA

honestadministrator
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X=30, using "area ACD", assume that AD=n, then we can solve it.

Andyyen-fqjh
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This solution is quite longer and harder.
Since, the ∆'s ADB and ADC lie on base ( d is midpoint) .
So, their area is equal.
We know area of ∆ ADB = 1/2 × AD×AB× sin(45°)
=>Area of ∆ ADC = 1/2 × AD × AC × sin(x)
=> 1/2×AD×AB = 1/2×AD×AC
From here we can solve further to get that x = 30°.

arulbiswas
Автор

Καλημέρα. Σήμερα είδα και έλυσα την άσκηση με τον παρακάτω τρόπο. Προεκτείνω την AD κατά τμήμα DE=AD. Tότε το ABEC είναι παραλληλόγραμμο (οι διαγώνιοί του διχοτομούνται). Άρα: ΒΕ=AC και <ΒΕΑ=x ως εντός εναλλάξ. Από νόμο ημιτόνων στο ΑΒΕ:

ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μμ
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Area of those 2 triangles are equals. same base, same height.
Area = ½ . S₁ . S₂ sin(α)
Area = ½ . S₂ . S₃. sin(β)
Equaling:
S₁ sin α = S₃ sin β
sin β = S₁ sin α / S₃
sin β = √3 sin 45° / √6
sin β = 0, 5
β = 30° ( Solved √ ) = Angle X

marioalb
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Dans un repère orthonorme de centre A d’axe AB l’homotetie de centre B derapport 2 envoie la droite AD vers la droite contenant C dont il suffit de trouver le point qui est à racine de 6 de A

LELULULU
Автор

Използвай формула за лице на триъгълник S=1/2*√3*AD*sin 45°, S=1/2*√6*AD*sin x°

xsilata