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

Solution:

Let the height of the vertical pole be AB and the point where the rope touches the ground be C. The length of the rope is AC, which is given as 20 m. The angle made by the rope with the ground level is ∠ACB = 30°.

We need to find the height of the pole, AB.

In the right-angled triangle ΔABC, we can use the sine trigonometric ratio, which relates the opposite side (height of the pole) to the hypotenuse (length of the rope).

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

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 rearrange the equation:

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

$$AB\; =\; 10$$

Therefore, the height of the pole is 10 m.

Solution:

Let the original height of the tree be AC. Suppose the tree breaks at point B due to a storm, and the broken part BD bends and touches the ground at point D. Let the foot of the tree be C.

The broken part BD is now AD (where A is the original top of the tree). The angle made by the broken part with the ground is ∠ADC = 30°.

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

We need to find the original height of the tree, which is AC = AB + BC. Note that BC is the part of the tree that did not break, and BD is the broken part which is equal to AD.

In the right-angled triangle ΔBCD (where the angle at B is 90°), we have:

1. To find the height of the standing part (BC):

We use the tangent ratio, which relates the opposite side (BC) to the adjacent side (CD).

$$\backslash tan(\backslash angle\; ADC)\; =\; \backslash frac\{\backslash text\{Opposite\}\}\{\backslash text\{Adjacent\}\}\; =\; \backslash frac\{BC\}\{CD\}$$

Substituting the given values:

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

We know that $\backslash tan(30^\{\backslash circ\})\; =\; \backslash frac\{1\}\{\backslash sqrt\{3\}\}$.

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

Solving for BC:

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

2. To find the length of the broken part (AD, which is equal to BD):

We use the cosine ratio, which relates the adjacent side (CD) to the hypotenuse (BD).

$$\backslash cos(\backslash angle\; ADC)\; =\; \backslash frac\{\backslash text\{Adjacent\}\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{CD\}\{BD\}$$

Substituting the given values:

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

We know that $\backslash cos(30^\{\backslash circ\})\; =\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$.

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

Solving for BD:

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

Since AD = BD, the length of the broken part is $\backslash frac\{16\}\{\backslash sqrt\{3\}\}$ m.

3. Calculate the total height of the tree:

The original height of the tree is the sum of the standing part (BC) and the broken part (BD).

$$\backslash text\{Height\; of\; the\; tree\}\; =\; BC\; +\; BD$$

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

To simplify, we can rationalize the denominator:

$$\backslash text\{Height\; of\; the\; tree\}\; =\; \backslash frac\{24\}\{\backslash sqrt\{3\}\}\; \backslash times\; \backslash frac\{\backslash sqrt\{3\}\}\{\backslash sqrt\{3\}\}\; =\; \backslash frac\{24\backslash sqrt\{3\}\}\{3\}\; =\; 8\backslash sqrt\{3\}\; \backslash text\{\; m\}$$

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