In the figure- the - ABC -30- - If CB -18 m- then AC -A- 9 -3 mB- 9 mC- 3 m3- 3 -3 p

The correct option is B 9 mGiven, ∠ ABC = 45° and BC = 18 m. nbsp; We know that, sin 30° = 1/2 From the figure, sin 30° = AC/BC ⇒1/2 = AC/18 ⇒ AC = 9 m ...

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Standard X

Mathematics

Ratio of Sides of Special Triangles

In the figure...

Question

In the figure, the $∠ A B C = 30 °$ . If $C B = 18 m$ , then AC = ________.

9 m

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9 \sqrt{3} m

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3 m

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3 \sqrt{3} m

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Solution

The correct option is A $9 m$
Given, $∠ A B C = 45 °$ and BC = 18 m.

We know that, $s i n 30 ° = \frac{1}{2}$

From the figure,
$s i n 30 ° = \frac{A C}{B C} \Rightarrow \frac{1}{2} = \frac{A C}{18} \Rightarrow A C = 9 m$

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In the figure, the ∠ABC=30°. If CB=18 m , then AC = ________.

The correct option is A 9 mGiven, ∠ABC=45° and BC = 18 m. We know that, sin 30°=12 From the figure, sin 30°=ACBC⇒12=AC18⇒AC=9 m

In the figure, the $∠ A B C = 30 °$ . If $C B = 18 m$ , then AC = ________.

The correct option is A $9 m$
Given, $∠ A B C = 45 °$ and BC = 18 m.

We know that, $s i n 30 ° = \frac{1}{2}$

From the figure,
$s i n 30 ° = \frac{A C}{B C} \Rightarrow \frac{1}{2} = \frac{A C}{18} \Rightarrow A C = 9 m$