In the adjoining figure, =90^ , = ,BC=CD=4cm and AD = 10cm. Find … - Vidyadip

Question

In the adjoining figure, $\angle\text{B}=90^\circ,\angle\text{BAC}=\theta,\text{BC}=\text{CD}=4\text{cm}$ and AD = 10cm. Find

$\sin\theta$
$\cos\theta.$

Hint: $\text{AB}^2=\big(\text{AD}^2-\text{BD}^2\big)=36\text{cm}^2$ $\therefore\text{AB}=6\text{cm}.$ $\text{AC}^2=\big(\text{AB}^2+\text{BC}^2\big)=52\text{cm}^2$ $\therefore\text{AC}=2\sqrt{13}\text{cm}.$ Thus, AB = 6cm and $\text{AC}=2\sqrt{13}\text{cm}.$

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Answer

In $\triangle\text{ABC},$ $\angle\text{B}=90^\circ$
$\text{AD}=10\text{cm}$
$\text{BD}=\text{BC}+\text{CD}$
$=4+4=8\text{cm}$
By pythagoras theoram, we have
$\text{AB}^2 = \text{AD}^2 - \text{BD}^2$
$=10^2-8^2=100-64=36$
$\Rightarrow\text{AB}=6\text{cm}$
Now, in $\triangle\text{ABC},\angle\text{B}=90^\circ.$
By Pythagoras theorem, we have
$\text{AC}^2 = \text{AB}^2 - \text{BC}^2$
$=6^2+4^2=36+16=52$
$\Rightarrow\text{AC}=2\sqrt{13}\text{cm}$

$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$=\frac{\text{BC}}{\text{AC}}=\frac{4}{2\sqrt{13}}=\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}$

$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$

$=\frac{\text{AB}}{\text{AC}}=\frac{6}{2\sqrt{13}}=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}.$

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