Question
In right angled $\triangle LMN$, if $\angle N =\theta, \angle M =90^{\circ}, \cos \theta=\frac{24}{25}$, find $\sin \theta$ and $\tan \theta$. Similarly, find $\left(\sin ^2 \theta\right)$ and $\left(\cos ^2 \theta\right)$
✓
Answer
$\text { i. } \cos \theta=\frac{24}{25}$
$\text { In } \Delta LMN , \angle M =90^{\circ}, \angle N =\theta$
$\therefore \quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \cos \theta=\frac{ MN }{ LN } \quad \ldots \text { (ii) }$
$\therefore \quad \frac{ MN }{ LN }=\frac{24}{25}$
Let the common multiple be k.
$\therefore MN = 24k$ and $LN = 25k$
Now, $LN^2= LM^2 + MN^2 … [Pythagoras theorem]$
$\therefore (25k)^2 = LM^2 + (24k)^2$
$\therefore 625 k^2 = LM^2 + 576k^2$
$\therefore LM^2 = 625k^2 – 576k^2$
$\therefore LM^2 = 49k^2$
$\therefore LM =\sqrt{49 k^2} \ldots \text {.TTaking square root of both sides] }$
$=7 k$
$\text { ii. } \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{L M}{L N}=\frac{7 k }{25 k }=\frac{7}{25}$
$\therefore \quad \sin ^2 \theta=(\sin \theta)^2=\left(\frac{7}{25}\right)^2=\frac{49}{625}$
iii. $\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{ LM }{ MN }=\frac{7 k }{24 k }=\frac{7}{24}$
iv. $\cos \theta=\frac{24}{25}$
...[Given]
$\therefore \quad \cos ^2 \theta=(\cos \theta)^2=\left(\frac{24}{25}\right)^2=\frac{576}{625}$
$\text { In } \Delta LMN , \angle M =90^{\circ}, \angle N =\theta$
$\therefore \quad \cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}$
$\therefore \quad \cos \theta=\frac{ MN }{ LN } \quad \ldots \text { (ii) }$
$\therefore \quad \frac{ MN }{ LN }=\frac{24}{25}$
Let the common multiple be k.
$\therefore MN = 24k$ and $LN = 25k$
Now, $LN^2= LM^2 + MN^2 … [Pythagoras theorem]$
$\therefore (25k)^2 = LM^2 + (24k)^2$
$\therefore 625 k^2 = LM^2 + 576k^2$
$\therefore LM^2 = 625k^2 – 576k^2$
$\therefore LM^2 = 49k^2$
$\therefore LM =\sqrt{49 k^2} \ldots \text {.TTaking square root of both sides] }$
$=7 k$
$\text { ii. } \sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{L M}{L N}=\frac{7 k }{25 k }=\frac{7}{25}$
$\therefore \quad \sin ^2 \theta=(\sin \theta)^2=\left(\frac{7}{25}\right)^2=\frac{49}{625}$
iii. $\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}=\frac{ LM }{ MN }=\frac{7 k }{24 k }=\frac{7}{24}$
iv. $\cos \theta=\frac{24}{25}$
...[Given]
$\therefore \quad \cos ^2 \theta=(\cos \theta)^2=\left(\frac{24}{25}\right)^2=\frac{576}{625}$
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