Question : $\triangle$LMN is right angled at M. If $\angle$N = 45$^\circ$, what is the length of MN (in cm), if NL = 9$\sqrt2$cm?
Option 1: $9\sqrt2$
Option 2: $\frac{9}{\sqrt2}$
Option 3: $18$
Option 4: $9$
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Correct Answer: $9$
Solution : $\triangle$LMN is right angled at M. $\angle$M = 90$^\circ$ $\angle$N = 45$^\circ$ $\angle$L = 180$^\circ$ – (45$^\circ$ + 90$^\circ$) = 45$^\circ$ So, $\triangle$LMN is an isosceles right-angled triangle. ⇒ LM = MN Also, NL = 9$\sqrt2$ cm Using the Pythagoras theorem, LM$^2$ + MN$^2$ = LN$^2$ ⇒ 2MN$^2$ = (9$\sqrt2$)$^2$ ⇒ MN = 9 cm Hence, the correct answer is $9$.
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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?
Question : In a right-angled triangle $\Delta PQR, PR$ is the hypotenuse of length 20 cm, $\angle PRQ = 30^{\circ}$, the area of the triangle is:
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