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Question Number 205535 by cortano12 last updated on 24/Mar/24

	Answered by A5T last updated on 24/Mar/24

	Commented by A5T last updated on 24/Mar/24

	$\frac{s i n 90 = 1}{B C} = \frac{s i n (90 - θ) = c o s θ}{A B} \Rightarrow A B = B C c o s θ$ $\frac{s i n 90 = 1}{B C} = \frac{s i n θ}{A C} \Rightarrow A C = B C s i n θ$ $∠ D B A = 180 - θ; ∠ A C E = 90 + θ$ $\Rightarrow {A D}^{2} = x^{2} {B C}^{2} + {B C}^{2} {c o s}^{2} θ + 2 {x B C}^{2} {c o s}^{2} θ$ $\Rightarrow {A E}^{2} = {B C}^{2} {s i n}^{2} θ + x^{2} {B C}^{2} + 2 {x B C}^{2} {s i n}^{2} θ$ ${D E}^{2} = {(2 x B C + B C)}^{2} = {B C}^{2} {(2 x + 1)}^{2}$ $\Rightarrow \frac{{A D}^{2} + {D E}^{2} + {E A}^{2}}{{B C}^{2}} = 2 x^{2} + 2 x + 1 + {(2 x + 1)}^{2}$ $\Rightarrow f (x) = 6 x^{2} + 6 x + 2 \Rightarrow f (\frac{\sqrt{1347} - 1}{2}) = 2021$
	Answered by mr W last updated on 24/Mar/24

	Commented by mr W last updated on 24/Mar/24

	${A D}^{2} = {(x + 1)}^{2} a^{2} + q^{2} - 2 a (x + 1) q \times \frac{q}{a}$ $= {(x + 1)}^{2} a^{2} - q^{2} (2 x + 1)$ $s i m i l a r l y$ ${A E}^{2} = {(x + 1)}^{2} a^{2} - p^{2} (2 x + 1)$ ${A D}^{2} + {A E}^{2} = 2 {(x + 1)}^{2} a^{2} - (p^{2} + q^{2}) (2 x + 1)$ $= 2 {(x + 1)}^{2} a^{2} - a^{2} (2 x + 1)$ $= (2 x^{2} + 2 x + 1) a^{2}$ ${D E}^{2} = {(2 x + 1)}^{2} a^{2} = (4 x^{2} + 4 x + 1) a^{2}$ $\Rightarrow f (x) = \frac{(2 x^{2} + 2 x + 1) a^{2} + (4 x^{2} + 4 x + 1) a^{2}}{a^{2}}$ $= 6 x^{2} + 6 x + 2 = 6 {(x + \frac{1}{2})}^{2} + \frac{1}{2}$ $f (\frac{\sqrt{1349} - 1}{2}) = 6 {(\frac{\sqrt{1349}}{2})}^{2} + \frac{1}{2}$ $= \frac{3 \times 1349 + 1}{2} = 2024 ✓$
	Commented by cortano12 last updated on 24/Mar/24


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