Question

In the adjoining figure triangle ABC is right angled at B if angle A=30 degree and AB = 9cm then find BC , AC

Answer 1

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Answer 2

BC = $3\sqrt{3}\ cm$

AC =  $6 \sqrt{3} \ cm$

Step-by-step explanation:

Given:- In $\triangle ABC$,

$\angle A= 30\ degree$, AB = 9 cm.

To find:- BC and AC.

Solution:- In $\triangle ABC$,

$tan\theta=\frac{opposite}{adjacent}$

$tan30=\frac{BC}{9}$    --------------------(given)

$\frac{1}{\sqrt{3} }=\frac{BC}{9}$

$BC=\frac{9}{\sqrt{3} }$

$BC=\frac{\sqrt{3}\times\sqrt{3}\times3 }{\sqrt{3} }$    -------------(∵$9=\sqrt{3}\times \sqrt{3}\times\ 3$)

$\therefore BC= 3\sqrt{3}\ cm$    -----------(equation 1)

Now,

$cos\theta=\frac{adjacent}{hypotenuse}$

$cos30=\frac{AB}{AC}$     ------------------(given $\theta =30\ degree$)

$\frac{\sqrt{3} }{2} =\frac{9}{AC}$         -----------------($cos 30=\frac{\sqrt{3} }{2}$  and AB=9 cm)

$AC=\frac{9\times 2}{\sqrt{3} }$

$\therefore$ AC = $\frac{\sqrt{3}\times \sqrt{3} \times 3\times 2 }{\sqrt{3} }$    ----------(∵$9=\sqrt{3}\times \sqrt{3}\times\ 3$)

$\therefore AC =6\sqrt{3}\ cm$