In right angled $∆$ LMN, $∠$ LMN = 90^° $∠$ L = 50^° and $∠$ N = 40^°, write the following ratios.

(i) sin 50^°(ii) cos 50^°(iii) tan 40^° (iv) cos 40^°

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Solution

(i) sin50º = $\frac{Opposite side of ∠ L}{Hypotenuse} = \frac{MN}{LN}$

(ii) cos50º = $\frac{Adjacent side of ∠ L}{Hypotenuse} = \frac{LM}{LN}$

(iii) tan40º = $\frac{Opposite side of ∠ N}{Adjacent side of ∠ N} = \frac{LM}{MN}$

(iv) cos40º = $\frac{Adjacent side of ∠ N}{Hypotenuse} = \frac{MN}{LN}$

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∣ ∣ ∣ \frac{sin 50}{cos 40} + \frac{csc 40}{sec 50} - 4 cos 50 x csc 40 ∣ ∣ ∣

∠

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