Using the information in the given figure- AC - m-A- 15-3-1B- 15-3-1C- 15D- 30

The correct option is B 15 (√(3) + 1) Let AC = x m Δ ACD is isosceles ∴ DC = AC = x In Δ ABC, tan 30 =x/30 + x 1/√(3) =x/30 + x 30 + x = x √(3) 30 = x√(3) x ...

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Standard IX

Mathematics

Values of Trigonometric Ratios

Using the inf...

Question

Using the information in the given figure, AC = ___ m.

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$15 (\sqrt{3} - 1)$

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$15 (\sqrt{3} + 1)$

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Solution

The correct option is D
$15 (\sqrt{3} + 1)$

Let AC = x m

$Δ A C D$ is isosceles

$∴$ DC = AC = x

In $Δ A B C, t a n 30 = \frac{x}{30 + x}$

$\frac{1}{\sqrt{3}} = \frac{x}{30 + x}$

$30 + x = x \sqrt{3}$

$30 = x \sqrt{3} - x$

$30 = x (\sqrt{3} - 1)$

$x = \frac{30}{\sqrt{3} - 1}$

$= \frac{30}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$= \frac{30 (\sqrt{3} + 1)}{2}$

$= 15 (\sqrt{3} + 1) m$

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Using the information in the given figure, AC = ___ m.

The correct option is D 15(√3+1) Let AC = x m ΔACD is isosceles ∴ DC = AC = x In ΔABC, tan 30=x30+x 1√3=x30+x 30+x=x√3 30=x√3−x 30=x(√3−1) x=30√3−1 =30√3−1×√3+1√3+1 =30(√3+1)2 =15(√3+1)m