Question : In $\triangle$LMN, LM = $5\sqrt{2}$ cm, LN = 13 cm and $\angle$LMN=135$^{\circ}$. What is the length (in cm) of MN?
Option 1: $7$
Option 2: $8$
Option 3: $8\sqrt{2}$
Option 4: $7\sqrt{2}$
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Correct Answer: $7$
Solution :
Use: cos 135º = $\frac{b^{2} + c^{2} - a^{2}}{2bc}$ where $b$ and $c$ are the sides of angle whereas $a$ is the side opposite to the angle.
According to the question
$a$ = 13 cm, $b$ = $5\sqrt{2}$ cm and let $c$ = $x$ cm
⇒ cos 135º = $\frac{b^{2} + c^{2} - a^{2}}{2bc}$
⇒ cos(90º + 45º) = $\frac{b^{2} + c^{2} - a^{2}}{2bc}$
⇒ – sin 45º = $\frac{b^{2} + c^{2} - a^{2}}{2×5\sqrt{2}× x }$
⇒ $\frac{-1}{\sqrt{2}}$ = $\frac{50 + x^{2} - 169}{10\sqrt{2}x}$
⇒ $–1 = \frac{x^{2} - 119}{10x}$
⇒ x$^{2}$ + 10$x$ – 119 = 0
⇒ $x(x - 7)(x + 17)$ = 0
⇒ $x$ = 7 cm [since side can not be negative.]
Hence, the correct answer is 7.
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