Question

In $\Delta ABC$, $AB=1$, $BC=1$ and $AC=1$ $\sqrt{2}$. In $\Delta MNP$, $MN=1$, $NP=1$ and $\angle$ $MNP=2$ $\angle$ $ABC$

The side $MP$ equals?

• A. $3\sqrt{2}$
• B. $7/4$
• C. $2\sqrt{2}$
• D. $\sqrt{7}/2$

Answer 1

The correct option is D $\sqrt{7}/2$

Consider the given triangles.
Applying cosine rule to $△ABC$ we get
$cos\left(\theta \right)=\frac{a^2+c^2-b^2}{2ac}$
$=\frac{2-\frac{1}{2}}{2}$
$=\frac{3}{4}$
Now
$cos2\theta =2co{s}^2\theta -1$
$=(2\times \frac{9}{16})-1$
$=\frac{1}{8}$
Now applying cosine rule to the $△MNP$ we get
$cos2\theta =\frac{m^2+p^2-n^2}{2mp}$
$\frac{1}{8}=\frac{2-n^2}{2}$
Or
$2-n^2=\frac{1}{4}$
Or
$n^2=\frac{7}{4}$
$n=\frac{\sqrt{7}}{2}$