Question

Find the length of EF.​

Answer 1

$\Huge\text{ {\overline{EF}=(10\sqrt{2}-5\sqrt{6}) \text{ cm}} }$

$\huge\text{\underline{\underline{Topic}}}$

$\Large\text{[Trigonometry]}$

$\large\text{ \rightarrow Double angle formula}$

$\text{ \bullet\ \boxed{\sin(2\theta)=2\sin\theta\cos\theta} \dots [Sine]}$

$\text{ \bullet\ \boxed{\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta} \dots[Cosine]}$

$\text{ \bullet\ \boxed{\tan(2\theta)=\dfrac{2\tan\theta}{1-\tan^{2}\theta}} \dots[Tangent]}$

$\huge\text{\underline{\underline{Explanation}}}$

Let the midpoint on segment BC be M.

Since $\triangle BCD$ is an equilateral right triangle, -

$\cdots\longrightarrow\angle BCD=45^{\circ}$

Since $\text{ \tan\angle CDM=\dfrac{1}{\sqrt{3}} }$, -

$\cdots\longrightarrow\angle CDM=30^{\circ}$

We know that -

$\bullet\ \angle BDC=45^{\circ}$

$\bullet\ \angle CDM=30^{\circ}$

$\cdots\longrightarrow\angle BDM=\angle BDC-\angle CDM=15^{\circ}$

Since the two diagonal of square cross perpendicularly, -

$\cdots\longrightarrow\angle E=90^{\circ}$

We know that -

$\bullet\ \overline{DE}=5\sqrt{2}$

$\bullet\ \angle E=90^{\circ}$

$\bullet\ \angle EDF=15^{\circ}$

$\cdots\longrightarrow\overline{EF}=5\sqrt{2}\tan15^{\circ}$

By double angle formula for tangent, -

$\text{ \cdots\longrightarrow\tan30^{\circ}=\dfrac{2\tan15^{\circ}}{1-\tan15^{\circ}} }$

$\bold{Let}\tan15^{\circ}\bold{be}\text{ t .}$

$\text{ \cdots\longrightarrow\dfrac{1}{\sqrt{3}}=\dfrac{2t}{1-t^{2}} }$

$\cdots\longrightarrow1-t^{2}=2\sqrt{3}t$

$\cdots\longrightarrow t^{2}+2\sqrt{3}t-1=0$

$\cdots\longrightarrow t=-\sqrt{3}\pm2,\ \boxed{t>0}$

$\cdots\longrightarrow\tan15^{\circ}=2-\sqrt{3}$

Now, finally, we know that -

$\bullet\ \overline{EF}=5\sqrt{2}\tan15^{\circ}$

$\bullet\ \tan15^{\circ}=2-\sqrt{3}$

$\text{ \cdots\longrightarrow\overline{EF}=10\sqrt{2}-5\sqrt{6} (cm)}$

$\huge\text{\underline{\underline{Final answer}}}$

$\text{ \cdots\longrightarrow\boxed{\overline{EF}=(10\sqrt{2}-5\sqrt{6}) \text{ cm}} }$