Three Areas as an Angle Vanishes
Difficulty: Hard · Topics: Calculus, Limits, Trigonometry
Burchak yo‘qolganda uchta yuza
In the triangle ABC we have angle ABC=2θ and angle ACB=3θ, and H is the foot of the perpendicular from the vertex A to the side BC. On the other side of BC from A there is a right triangle CBD having BC as its hypotenuse, with angle CBD=θ and the right angle at D.
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Let S 1, S 2, S 3 be the areas of the triangles ABH, AHC, CBD respectively. Find the value of lim θ→ 0+fracS 1+S 3S 2.
ABC uchburchakda angle ABC=2θ va angle ACB=3θ, H — A uchidan BC tomonga tushirilgan perpendikulyar asosi. BC ning A ga qarama-qarshi tomonida gipotenuzasi BC bo‘lgan, angle CBD=θ va to‘g‘ri burchagi D da bo‘lgan CBD to‘g‘ri burchakli uchburchak bor.
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S 1, S 2, S 3 — mos ravishda ABH, AHC, CBD uchburchaklar yuzalari bo‘lsin. lim θ→ 0+fracS 1+S 3S 2 ning qiymatini toping.