CBSE Class 10 Mathematics Chapter 9: Some Applications of Trigonometry NCERT Solutions

Let AB represent the vertical pole and AC be the rope. The length of the rope (AC) is given as 20 m, and the angle made by the rope with the ground (angle ACB) is 30°.

We need to find the height of the pole, which is the length of AB.

In the right-angled triangle ABC, we can use the sine trigonometric ratio, which is defined as the ratio of the length of the side opposite to the angle to the length of the hypotenuse.

$$\backslash sin(\backslash angle\; ACB)\; =\; \backslash frac\{\backslash text\{Opposite\; side\}\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{AB\}\{AC\}$$

Substituting the given values:

$$\backslash sin(30^\{\backslash circ\})\; =\; \backslash frac\{AB\}\{20\}$$

We know that $\backslash sin(30^\{\backslash circ\})\; =\; \backslash frac\{1\}\{2\}$.

$$\backslash frac\{1\}\{2\}\; =\; \backslash frac\{AB\}\{20\}$$

To find AB, we can multiply both sides by 20:

$$AB\; =\; 20\; \backslash times\; \backslash frac\{1\}\{2\}$$

$$AB\; =\; 10$$

Therefore, the height of the pole is 10 meters.

Let the original height of the tree be AC. Suppose the tree breaks at point B due to a storm, and the broken part BC bends and touches the ground at point D, making an angle of 30° with the ground. The distance between the foot of the tree (A) and the point where the top touches the ground (D) is AD = 8 m.

In the right-angled triangle ABD, we have:

• The angle $\backslash angle\; ADB\; =\; 30^\{\backslash circ\}$.
• The adjacent side to this angle is AD = 8 m.
• The opposite side is AB (the broken part that is now standing).
• The hypotenuse is BD (the broken part that touches the ground).

First, let's find the length of the standing part AB using the tangent ratio:

$$\backslash tan(\backslash angle\; ADB)\; =\; \backslash frac\{\backslash text\{Opposite\; side\}\}\{\backslash text\{Adjacent\; side\}\}\; =\; \backslash frac\{AB\}\{AD\}$$

$$\backslash tan(30^\{\backslash circ\})\; =\; \backslash frac\{AB\}\{8\}$$

Since $\backslash tan(30^\{\backslash circ\})\; =\; \backslash frac\{1\}\{\backslash sqrt\{3\}\}$:

$$\backslash frac\{1\}\{\backslash sqrt\{3\}\}\; =\; \backslash frac\{AB\}\{8\}$$

Solving for AB:

$$AB\; =\; \backslash frac\{8\}\{\backslash sqrt\{3\}\}\; \backslash text\{\; m\}$$

Next, let's find the length of the broken part BD using the cosine ratio:

$$\backslash cos(\backslash angle\; ADB)\; =\; \backslash frac\{\backslash text\{Adjacent\; side\}\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{AD\}\{BD\}$$

$$\backslash cos(30^\{\backslash circ\})\; =\; \backslash frac\{8\}\{BD\}$$

Since $\backslash cos(30^\{\backslash circ\})\; =\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$:

$$\backslash frac\{\backslash sqrt\{3\}\}\{2\}\; =\; \backslash frac\{8\}\{BD\}$$

Solving for BD:

$$BD\; =\; \backslash frac\{8\; \backslash times\; 2\}\{\backslash sqrt\{3\}\}\; =\; \backslash frac\{16\}\{\backslash sqrt\{3\}\}\; \backslash text\{\; m\}$$

The original height of the tree is the sum of the standing part (AB) and the broken part (BD):

$$\backslash text\{Original\; height\}\; =\; AB\; +\; BD\; =\; \backslash frac\{8\}\{\backslash sqrt\{3\}\}\; +\; \backslash frac\{16\}\{\backslash sqrt\{3\}\}$$

$$\backslash text\{Original\; height\}\; =\; \backslash frac\{8\; +\; 16\}\{\backslash sqrt\{3\}\}\; =\; \backslash frac\{24\}\{\backslash sqrt\{3\}\}\; \backslash text\{\; m\}$$

To rationalize the denominator, multiply the numerator and denominator by $\backslash sqrt\{3\}$:

$$\backslash text\{Original\; height\}\; =\; \backslash frac\{24\}\{\backslash sqrt\{3\}\}\; \backslash times\; \backslash frac\{\backslash sqrt\{3\}\}\{\backslash sqrt\{3\}\}\; =\; \backslash frac\{24\backslash sqrt\{3\}\}\{3\}\; =\; 8\backslash sqrt\{3\}\; \backslash text\{\; m\}$$

Hence, the height of the tree is $8\backslash sqrt\{3\}$ m.