NCERT Solutions: Class 10 Maths Chapter 9 - Some Applications of Trigonometry

The rope is the hypotenuse of length 20 m, and the pole is opposite the $30^\circ$ angle. Thus $\sin30^\circ=\dfrac{h}{20}$, so $h=20\times\dfrac12=10$ m.

Let the unbroken vertical part be $h$ and the broken part be $l$. The horizontal distance is 8 m and the broken part makes $30^\circ$ with the ground. So $h=8\tan30^\circ=\dfrac{8}{\sqrt3}$ and $l=\dfrac{8}{\cos30^\circ}=\dfrac{16}{\sqrt3}$. Total height $=h+l=\dfrac{24}{\sqrt3}=8\sqrt3$ m.

Slide length is the hypotenuse. For the smaller slide, $\sin30^\circ=\dfrac{1.5}{l}$, so $l=3$ m. For the steeper slide, $\sin60^\circ=\dfrac{3}{l}$, so $l=\dfrac{3}{\sqrt3/2}=2\sqrt3$ m.

$\tan30^\circ=\dfrac{h}{30}$, so $h=30\times\dfrac1{\sqrt3}=10\sqrt3$ m.

If the string length is $l$, then $\sin60^\circ=\dfrac{60}{l}$. Thus $l=\dfrac{60}{\sqrt3/2}=40\sqrt3$ m.

The vertical height above his eyes is $30-1.5=28.5$ m. If the initial and final horizontal distances are $x$ and $y$, then $\tan30^\circ=\dfrac{28.5}{x}$, so $x=28.5\sqrt3$, and $\tan60^\circ=\dfrac{28.5}{y}$, so $y=\dfrac{28.5}{\sqrt3}=9.5\sqrt3$. Distance walked $=x-y=19\sqrt3$ m.

Let the horizontal distance be $x$. The angle to the top of the 20 m building is $45^\circ$, so $\tan45^\circ=\dfrac{20}{x}$ and $x=20$. If total height up to the top of the tower is $H$, then $\tan60^\circ=\dfrac{H}{20}$, so $H=20\sqrt3$. Tower height $=20\sqrt3-20=20(\sqrt3-1)$ m.

Let the pedestal height be $h$ and the horizontal distance be $x$. From the top of the pedestal, $\tan45^\circ=\dfrac{h}{x}$, so $x=h$. From the top of the statue, $\tan60^\circ=\dfrac{h+1.6}{x}=\dfrac{h+1.6}{h}$. Thus $h\sqrt3=h+1.6$, so $h=\dfrac{1.6}{\sqrt3-1}=0.8(\sqrt3+1)$ m.

Let the distance between tower and building be $d$, and building height be $h$. From the foot of the building to the top of the 50 m tower, $\tan60^\circ=\dfrac{50}{d}$, so $d=\dfrac{50}{\sqrt3}$. From the foot of the tower to the top of the building, $\tan30^\circ=\dfrac{h}{d}$, so $h=\dfrac{d}{\sqrt3}=\dfrac{50}{3}$ m.

Let the distances from the point to the two poles be $x$ and $y$, with $x+y=80$, and let the common height be $h$. Then $\tan60^\circ=\dfrac{h}{x}$, so $h=\sqrt3x$, and $\tan30^\circ=\dfrac{h}{y}$, so $h=\dfrac{y}{\sqrt3}$. Hence $y=3x$, so $x=20$, $y=60$, and $h=20\sqrt3$ m.

Let the width of the canal be $x$ and tower height be $h$. From the opposite bank, $\tan60^\circ=\dfrac{h}{x}$, so $h=\sqrt3x$. From the point 20 m farther away, $\tan30^\circ=\dfrac{h}{x+20}$, so $h=\dfrac{x+20}{\sqrt3}$. Equating gives $3x=x+20$, so $x=10$ and $h=10\sqrt3$.

The angle of depression to the tower foot is $45^\circ$, so the horizontal distance equals 7 m. The angle of elevation from the building top to the tower top is $60^\circ$, so the extra height above the building top is $7\tan60^\circ=7\sqrt3$ m. Therefore tower height $=7+7\sqrt3=7(\sqrt3+1)$ m.

Angles of depression equal angles of elevation. The nearer ship at $45^\circ$ is 75 m from the lighthouse foot. The farther ship at $30^\circ$ is $\dfrac{75}{\tan30^\circ}=75\sqrt3$ m away. Their distance apart is $75\sqrt3-75=75(\sqrt3-1)$ m.

The vertical height above the girl's eyes is $88.2-1.2=87$ m. At $60^\circ$, the horizontal distance is $\dfrac{87}{\tan60^\circ}=29\sqrt3$ m. At $30^\circ$, it is $\dfrac{87}{\tan30^\circ}=87\sqrt3$ m. Distance travelled horizontally $=87\sqrt3-29\sqrt3=58\sqrt3$ m.

Let the tower height be $h$. When the angle of depression is $30^\circ$, the car's distance from the foot is $h\cot30^\circ=\sqrt3h$. Six seconds later, at $60^\circ$, the distance is $h\cot60^\circ=\dfrac{h}{\sqrt3}$. In 6 seconds it covers $\sqrt3h-\dfrac{h}{\sqrt3}=\dfrac{2h}{\sqrt3}$. The remaining distance is $\dfrac{h}{\sqrt3}$, half of that, so the remaining time is 3 seconds.