NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Ans : \( \begin{array}{l} \text{It can be observed from the figure that AB is the pole.}{\text { In } \triangle A B C,} \\ {\frac{A B}{A C}=\sin 30^{\circ}} \\ {\frac{A B}{20}=\frac{1}{2}} \\ {A B=\frac{20}{2}=10} \\ {\text { Therefore, the height of the pole is } 10 \mathrm{m} \text { . }}\end{array}\) 

Ans : 
Let AC was the original tree. Due to storm, it was broken into parts. The broken part  
\(\mathrm{A}^{\prime} \mathrm{B} \text { is making } 30^{\circ}\) with the ground.
\(\begin{array}{l}{\text { In } \Delta \mathrm{A}^{\prime} \mathrm{BC}} \\ {\frac{\mathrm{BC}}{\mathrm{A}^{\prime} \mathrm{C}}=\tan 30^{\circ}} \\ {\frac{\mathrm{BC}}{8}=\frac{1}{\sqrt{3}}} \\ {\mathrm{BC}=\left(\frac{8}{\sqrt{3}}\right) \mathrm{m}}\end{array}\)
\(\begin{array}{l}{\frac{\mathrm{A}^{\prime} \mathrm{C}}{\mathrm{A}^{\prime} \mathrm{B}}=\cos 30^{\circ}} \\ {\frac{8}{\mathrm{A}^{\prime} \mathrm{B}}=\frac{\sqrt{3}}{2}} \\ {\mathrm{A}^{\prime} \mathrm{B}=\left(\frac{16}{\sqrt{3}}\right) \mathrm{m}} \\ {\text { Height of tree }=\mathrm{A}^{\prime} \mathrm{B}+\mathrm{BC}}\end{array}\)
\(\begin{aligned} &=\left(\frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}}\right) \mathrm{m}=\frac{24}{\sqrt{3}} \mathrm{m} \\ &=8 \sqrt{3} \mathrm{m} \end{aligned}\)
Hence, the height of the tree is \(8 \sqrt{3} \mathrm{m}\). 

Ans : It can be observed that AC and PR are the slides for younger and elder children respectively.

\(\begin{array}{l}{\text { In } \triangle A B C,} \\ {\frac{A B}{A C}=\sin 30^{\circ}} \\ {\frac{1.5}{A C}=\frac{1}{2}} \\ {A C=3 \mathrm{m}}\end{array}\)

\(\begin{array}{l}{\text { In } \Delta \mathrm{PQR}_{,}} \\ {\frac{\mathrm{PQ}}{\mathrm{PR}}=\sin 60} \\ {\frac{3}{\mathrm{PR}}=\frac{\sqrt{3}}{2}} \\ {\mathrm{PR}=\frac{6}{\sqrt{3}}=2 \sqrt{3} \mathrm{m}}\end{array}\)
Therefore, the length of these slides are 3 m and \(2 \sqrt{3} \mathrm{m}\). 

Ans :  
Let AB be the tower and the angle of elevation from point C(on ground) is \(30^{\circ}\).
In \(\triangle A B C\),
\(\begin{array}{l}{\frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ}} \\ {\frac{\mathrm{AB}}{30}=\frac{1}{\sqrt{3}}} \\ {\mathrm{AB}=\frac{30}{\sqrt{3}}=10 \sqrt{3} \mathrm{m}}\end{array}\)
Therefore, the height of the tower is \(10 \sqrt{3} \mathrm{m}\). 

Ans : 
Let K be the kite and the string is tied to point P on the ground.
\(\begin{array}{l}{\text { In } \Delta \mathrm{KLP}_{,}} \\ {\frac{\mathrm{KL}}{\mathrm{KP}}=\sin 60^{\circ}} \\ {\frac{60}{\mathrm{KP}}=\frac{\sqrt{3}}{2}} \\ {\mathrm{KP}=\frac{120}{\sqrt{3}}=40 \sqrt{3} \mathrm{m}} \\ {\text { Hence, the length of the string is } 40 \sqrt{3} \mathrm{m} \text { . }}\end{array}\)