Question

In right angled $∆$ LMN , if $\angle$N = $\mathrm{\theta }$ , $\angle$M = ${90}^{°}$ , $\cos \mathrm{\theta }=\frac{24}{25}$, find sin $\mathrm{\theta }$ and tan $\mathrm{\theta }$ Similarly, find ( sin² $\mathrm{\theta }$) and ( cos² $\mathrm{\theta }$ ).

Answer 1

In right ∆LMN, ∠N = θ.

$$\begin{gathered} \cos \theta =\frac{24}{25} \\ \Rightarrow \frac{\mathrm{MN}}{\mathrm{LN}}=\frac{24}{25} \end{gathered}$$
Let MN = 24k and LN = 25k.

Using Pythagoras theorem, we have

LN² = LM² + MN²

⇒ (25k)² = LM² + (24k)²

⇒ LM² = 625k² − 576k² = 49k²

⇒ LM² = (7k)²

⇒ LM = 7k

Now,

$\sin \theta =\frac{\mathrm{LM}}{\mathrm{LN}}=\frac{7k}{25k}=\frac{7}{25}$

$\tan \theta =\frac{\mathrm{LM}}{\mathrm{NM}}=\frac{7k}{24k}=\frac{7}{24}$

Also,

${\sin }^2\theta ={\left(\frac{7}{25}\right)}^2=\frac{49}{625}$

${\cos }^2\theta ={\left(\frac{24}{25}\right)}^2=\frac{576}{625}$