mathematics , mensuration
Question-1: The lengths of the two sides of a triangle are $28\;\mathrm{m}$ , $26\;\mathrm{m}$ respectively and its are is $182\;\mathrm{m^2}$ . Find the other side of the triangle . Solution: $ \dfrac{1}{2} a b \sin \theta=182\mathrm{m^2}$ $\Rightarrow \dfrac{1}{2} \times 26 \times 28 \sin \theta=182$ $\Rightarrow 13 \times 28 \sin \theta=182$ $\Rightarrow \sin \theta=\dfrac{182}{13 \times 28}$ $\Rightarrow \sin \theta=0.5$ $\Rightarrow \theta=\sin ^{-1}(0.5)$ $\therefore \theta=30^{\circ}$ $ \cos \theta=\dfrac{b^{2}+a^{2}-c^{2}}{2 a b}$ $\Rightarrow \cos 30^{\circ}=\dfrac{26^{2}+28^{2}-c^{2}}{2 \times 26 \times 28} $ $\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{1460-c^{2}}{1456} $ $\Rightarrow \dfrac{1456 \sqrt{3}}{2}=1460-c^{2}$ $\Rightarrow 728 \sqrt{3}=1460-c^{2}$ $\Rightarrow c^2=1460-728 \sqrt{3}$ $\Rightarrow c=\sqrt{1460-728 \sqrt{3}}$ $\therefore c=14.11\;\mathrm{m^2}$ $\large{\textbf{Mathematical Questions:}}$ 1. Shakilal draws an equilateral triang...