CBSE Class 10 Maths Chapter 24: Some Applications of Trigonometry NCERT Solutions

Let AB represent the vertical pole and AC represent the rope. The length of the rope (AC) is given as 20 m. The angle made by the rope with the ground level (angle ACB) is given as 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 relates the opposite side (height of the pole, AB) to the hypotenuse (length of the rope, AC).

The formula is: $\backslash sin(\backslash theta)\; =\; \backslash frac\{\backslash text\{Opposite\}\}\{\backslash text\{Hypotenuse\}\}$

Substituting the given values:

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

We know that $\backslash sin(30^\{\backslash circ\})\; =\; \backslash frac\{1\}\{2\}$ and AC = 20 m.

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

To find AB, we can rearrange the equation:

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

$$AB\; =\; 10$$

Therefore, the height of the pole is 10 m.

Let the original height of the tree be AC. Due to a storm, the tree breaks at point B, and the broken part (BC) bends to touch the ground at point D. Thus, the original tree's height is AB + BC, and after breaking, the part BC becomes BD.

The top of the tree touches the ground at D, making an angle of 30° with the ground. So, $\backslash angle\; BDC\; =\; 30^\{\backslash circ\}$.

The distance between the foot of the tree (C) and the point where the top touches the ground (D) is given as 8 m. So, CD = 8 m.

In the right-angled triangle BCD:

We can find the length of the standing part of the tree (BC) using the tangent ratio:

$$\backslash tan(\backslash theta)\; =\; \backslash frac\{\backslash text\{Opposite\}\}\{\backslash text\{Adjacent\}\}$$

$$\backslash tan(30^\{\backslash circ\})\; =\; \backslash frac\{BC\}\{CD\}$$

Substituting the values, $\backslash tan(30^\{\backslash circ\})\; =\; \backslash frac\{1\}\{\backslash sqrt\{3\}\}$ and CD = 8 m:

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

Solving for BC:

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

Now, we can find the length of the broken part (BD) using the cosine ratio:

$$\backslash cos(\backslash theta)\; =\; \backslash frac\{\backslash text\{Adjacent\}\}\{\backslash text\{Hypotenuse\}\}$$

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

Substituting the values, $\backslash cos(30^\{\backslash circ\})\; =\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$ and CD = 8 m:

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

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 (BC) and the broken part (BD):

Height of the tree = $BC\; +\; BD$

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

To simplify, we can rationalize the denominator:

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

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