Question-172915

Question Number 172915 by mnjuly1970 last updated on 03/Jul/22

Commented by infinityaction last updated on 03/Jul/22

6 ?

Answered by mr W last updated on 03/Jul/22

Commented by mr W last updated on 03/Jul/22

\cos γ = \frac{4 - 1}{4 + 1} = \frac{3}{5}

{B D}^{2} = 4^{2} + 5^{2} + 2 \times 4 \times 5 \times \frac{3}{5} = 65

E B = \sqrt{{B D}^{2} - 1^{2}} = 8

\tan β = \frac{1}{8}

\tan α = \frac{4}{7}

\tan B = \tan (α + β) = \frac{\frac{4}{7} + \frac{1}{8}}{1 - \frac{4}{7} \times \frac{1}{8}} = \frac{39}{52}

A C = A B \times \tan B = 8 \times \frac{39}{52} = 6 ✓

Commented by Tawa11 last updated on 03/Jul/22

Great sir

Answered by som(math1967) last updated on 04/Jul/22

Commented by som(math1967) last updated on 04/Jul/22

l e t A C = x

∠ D C O = ∠ O C B = θ

∴ ∠ A C B = 2 θ

A E = 4 - 1 = 3, O E = \sqrt{5^{2} - 3^{2}} = 4

∴ C D = x - 4

t a n θ = \frac{1}{x - 4}, t a n 2 θ = \frac{8}{x}

t a n 2 θ = \frac{2 t a n θ}{1 - {t a n}^{2} θ}

\frac{8}{x} = \frac{\frac{2}{x - 4}}{\frac{{(x - 4)}^{2} - 1}{{(x - 4)}^{2}}}

\frac{8}{x} = \frac{2 (x - 4)}{x^{2} - 8 x + 15}

4 x^{2} - 32 x + 60 = x^{2} - 4 x

3 x^{2} - 28 x + 60 = 0

3 x^{2} - 18 x - 10 x + 60 = 0

(x - 6) (3 x - 10) = 0

(x - 6) = 0 [∵ x > 4 ∴ x > \frac{10}{3}, (3 x - 10) \neq 0]

∴ x = 6

Commented by Tawa11 last updated on 04/Jul/22

Great sir

Answered by infinityaction last updated on 04/Jul/22

Commented by infinityaction last updated on 04/Jul/22

D B = \sqrt{4^{2} + 7^{2}} = \sqrt{65}

△ D E B

B E = \sqrt{{(\sqrt{65})}^{2} - 1^{2}} = 8

△ A B C

{(x + 4)}^{2} + 8^{2} = {(x + 8)}^{2}

x^{2} + 16 + 8 x + 64 = x^{2} + 64 + 16 x

8 x = 16 \Rightarrow x = 2

A C = 4 + x = 4 + 2 = 6

Commented by Tawa11 last updated on 04/Jul/22

Great sir

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