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I tried to solve the given question in the following way. Where I am going wrong?

Q. Let $\vec A$ and $\vec B$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.

I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore\ \theta=60^\circ-30^\circ=30^\circ$
$\therefore\ |\vec R| = \sqrt{A^2+B^2+2AB \cos\theta}$
= $\sqrt{10^2+10^2+2\cdot10\cdot10\cos30^\circ}$
= $\sqrt{100+100+200\,\sqrt3/2}$
= $\sqrt{100+100+100\,\sqrt3}$
= $\sqrt{200+100\,\sqrt3}$
= $\sqrt{100\,(2+\sqrt3)}$
= $10\,\sqrt{2+\sqrt3}$

Now, I don't know how to solve it further. The answer is $20 \cos 15^\circ$. Please help me to simplify it further.

$$\cos 15^\circ = \frac{1+\sqrt3}{2\sqrt2}.$$

I tried to solve the given question in the following way. Where I am going wrong?

Q. Let $\vec A$ and $\vec B$ be the two vectors of magnitude $10$ unit each. If they are inclined to the $x$-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.

I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore\ \theta=60^\circ-30^\circ=30^\circ$
$\therefore\ |\vec R| = \sqrt{A^2+B^2+2AB \cos\theta}$
= $\sqrt{10^2+10^2+2\cdot10\cdot10\cos30^\circ}$
= $\sqrt{100+100+200\,\sqrt3/2}$
= $\sqrt{100+100+100\,\sqrt3}$
= $\sqrt{200+100\,\sqrt3}$
= $\sqrt{100\,(2+\sqrt3)}$
= $10\,\sqrt{2+\sqrt3}$

Now, I don't know how to solve it further. The answer is $20 \cos 15^\circ$. Please help me to simplify it further.

$$\cos 15^\circ = \frac{1+\sqrt3}{2\sqrt2}.$$

I tried to solve the given question in the following way. Where I am going wrong?

Q. Let $\vec A$ and $\vec B$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.

I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore$ $\theta$ = $60^\circ$ - $30^\circ$
= $30^\circ$
$\therefore$|$\overrightarrow R$| = $\sqrt{A^2+B^2+2AB cos\theta}$
= $\sqrt{10^2+10^2+2(10)(10)cos30^\circ}$
= $\sqrt{100+100+200(\frac{\sqrt3}2)}$
= $\sqrt{100+100+100{\sqrt3}}$
= $\sqrt{200+100{\sqrt3}}$
= $\sqrt{100(2+{\sqrt3})}$
= $10\sqrt{2+{\sqrt3}}$

Now, I don't know how to solve it further. The answer is $20 \cos 15^\circ$. Please help me to simplify it further.

$\cos 15^\circ = \frac{1+\sqrt3}{2\sqrt2}$

I tried to solve the given question in the following way. Where I am going wrong?

Q. Let $\vec A$ and $\vec B$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.

I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore\ \theta=60^\circ-30^\circ=30^\circ$
$\therefore\ |\vec R| = \sqrt{A^2+B^2+2AB \cos\theta}$
= $\sqrt{10^2+10^2+2\cdot10\cdot10\cos30^\circ}$
= $\sqrt{100+100+200\,\sqrt3/2}$
= $\sqrt{100+100+100\,\sqrt3}$
= $\sqrt{200+100\,\sqrt3}$
= $\sqrt{100\,(2+\sqrt3)}$
= $10\,\sqrt{2+\sqrt3}$

  Now, I don't know how to solve it further. The answer is $20 \cos 15^\circ$. Please help me to simplify it further.

$$\cos 15^\circ = \frac{1+\sqrt3}{2\sqrt2}.$$

I tried to solve the given question in the following way. Where I am going wrong?
Q. Let $\overrightarrow A$ and $\overrightarrow B$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.

I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore$ $\theta$ = $60^\circ$ - $30^\circ$
= $30^\circ$
$\therefore$|$\overrightarrow R$| = $\sqrt{A^2+B^2+2AB cos\theta}$
= $\sqrt{10^2+10^2+2(10)(10)cos30^\circ}$
= $\sqrt{100+100+200(\frac{\sqrt3}2)}$
= $\sqrt{100+100+100{\sqrt3}}$
= $\sqrt{200+100{\sqrt3}}$
= $\sqrt{100(2+{\sqrt3})}$
= $10\sqrt{2+{\sqrt3}}$

Now, I don't know how to solve it further. The answer is $20 cos 15^\circ$. Please help me to simplify it further.
$cos 15^\circ$ = $\frac{1+\sqrt3}{2\sqrt2}$

I tried to solve the given question in the following way. Where I am going wrong?

Q. Let $\vec A$ and $\vec B$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.

I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore$ $\theta$ = $60^\circ$ - $30^\circ$
= $30^\circ$
$\therefore$|$\overrightarrow R$| = $\sqrt{A^2+B^2+2AB cos\theta}$
= $\sqrt{10^2+10^2+2(10)(10)cos30^\circ}$
= $\sqrt{100+100+200(\frac{\sqrt3}2)}$
= $\sqrt{100+100+100{\sqrt3}}$
= $\sqrt{200+100{\sqrt3}}$
= $\sqrt{100(2+{\sqrt3})}$
= $10\sqrt{2+{\sqrt3}}$

Now, I don't know how to solve it further. The answer is $20 \cos 15^\circ$. Please help me to simplify it further.

$\cos 15^\circ = \frac{1+\sqrt3}{2\sqrt2}$