Question

A kite is flying with the string inclined at ${75}^o$ to the horizon. If the length of the string is $25m$, then height of the kite is :

• A. $\left(\frac{25}{2}\right){(\sqrt{3}-1)}^2$
• B. $\left(\frac{25}{2}\right)\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)$
• C. $\left(\frac{25}{2}\right){(\sqrt{3}+1)}^2$
• D. $\left(\frac{25}{2}\right)(\sqrt{6}+\sqrt{2})$

Answer 1

The correct option is B $\left(\frac{25}{2}\right)\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)$

Let H be the height of kite $\therefore \sin { 75 }^{ 0 }=\cfrac { H }{ 25 }

$$\sin { { 75 }^{ 0 } } =\sin { \left( { 45 }^{ 0 }+{ 30 }^{ ' } \right) } \sin\ 75^o=\sin(30^o+45^o)\\=\sin { { 45 }^{ 0 } } \cos { { 30 }^{ 0 } } +\cos { { 45 }^{ 0 } } \sin { { 30 }^{ 0 }
\\=\cfrac { 1 }{ \sqrt { 2 } } \times { \sqrt { 3 } }{ 2 } +\cfrac { 1 }{ \sqrt { 2 } } \times { 1 }{ 2 } }=\cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } $Putvalueof$\sin { { 75 }^{ 0 } } $inequation1$H=25\left[ \cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right] H=\cfrac { 25 }{ 2 } \left[ \cfrac { \sqrt { 3 } +1 }{ \sqrt { 2 } } \right] $