Some Applications Of TrigonometryClass 10 Mathematics NCERT Solutions

To Find: The height of the pole.

Solution: Let AB be the vertical pole and AC be the rope. The rope is stretched from the top of the pole A to a point C on the ground. This forms a right-angled triangle ABC, with the right angle at B.

Here, AC is the hypotenuse, AB is the side opposite to the angle at C (height of the pole), and BC is the side adjacent to the angle at C.

We have:

We can use the sine ratio, which relates the opposite side, hypotenuse, and the angle.

For $\triangle ABC$: $\sin 30^{\circ} = \frac{AB}{AC}$

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

$AB = \frac{20}{2}$ $AB = 10 \text{ m}$

Final Answer: The height of the pole is 10 m.

To Find: The original height of the tree.

Solution: Let the original tree be represented by the line segment AB. Suppose the tree breaks at point C. The broken part, CA, bends and its top A touches the ground at a point D. This forms a right-angled triangle BCD, with the right angle at B.

Here:

The original height of the tree is AB = AC + BC = CD + BC.

First, we find the length of BC (opposite side) using the tangent ratio in $\triangle BCD$: $\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}}$ $\tan 30^{\circ} = \frac{BC}{BD}$ $\frac{1}{\sqrt{3}} = \frac{BC}{8}$ $BC = \frac{8}{\sqrt{3}} \text{ m}$

Next, we find the length of CD (hypotenuse) using the cosine ratio in $\triangle BCD$: $\cos \theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}}$ $\cos 30^{\circ} = \frac{BD}{CD}$ $\frac{\sqrt{3}}{2} = \frac{8}{CD}$ $CD = \frac{8 \times 2}{\sqrt{3}} = \frac{16}{\sqrt{3}} \text{ m}$

The total height of the tree is the sum of the remaining part (BC) and the broken part (CD). Height of tree = BC + CD $= \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}}$ $= \frac{8+16}{\sqrt{3}} = \frac{24}{\sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $= \frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m}$

Final Answer: The height of the tree is $8\sqrt{3}$ m.

Given: Two scenarios for installing slides.

To Find: The length of the slide in each case.

Solution: In both cases, the slide, the vertical height, and the ground form a right-angled triangle. The length of the slide is the hypotenuse.

Case 1: For children below 5 years Let the height be AB = 1.5 m and the length of the slide be AC. The angle with the ground is $\angle ACB = 30^{\circ}$.

In right-angled $\triangle ABC$: We use the sine ratio: $\sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}}$ $\sin 30^{\circ} = \frac{AB}{AC}$ $\frac{1}{2} = \frac{1.5}{AC}$ $AC = 1.5 \times 2 = 3 \text{ m}$

Case 2: For elder children Let the height be PQ = 3 m and the length of the slide be PR. The angle with the ground is $\angle PRQ = 60^{\circ}$.

In right-angled $\triangle PQR$: We use the sine ratio: $\sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}}$ $\sin 60^{\circ} = \frac{PQ}{PR}$ $\frac{\sqrt{3}}{2} = \frac{3}{PR}$ $PR = \frac{3 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}}$

To rationalize the denominator: $PR = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \text{ m}$

Final Answer: The length of the slide for younger children should be 3 m, and the length of the slide for elder children should be $2\sqrt{3}$ m.

To Find: The height of the tower.

Solution: Let AB be the tower and C be the point on the ground. This forms a right-angled triangle ABC, with the right angle at B.

Here:

We can use the tangent ratio, which relates the opposite side, adjacent side, and the angle.

For $\triangle ABC$: $\tan 30^{\circ} = \frac{AB}{BC}$

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

$\frac{1}{\sqrt{3}} = \frac{AB}{30}$ $AB = \frac{30}{\sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $AB = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$

Final Answer: The height of the tower is $10\sqrt{3}$ m.

To Find: The length of the string.

Solution: Let A be the position of the kite and C be the point on the ground where the string is tied. Let B be the point on the ground directly below the kite. This forms a right-angled triangle ABC, with the right angle at B.

Here:

We can use the sine ratio, which relates the opposite side, hypotenuse, and the angle.

For $\triangle ABC$: $\sin 60^{\circ} = \frac{AB}{AC}$

We know that $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.

$\frac{\sqrt{3}}{2} = \frac{60}{AC}$ $AC = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $AC = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \text{ m}$

Final Answer: The length of the string is $40\sqrt{3}$ m.

To Find: The distance the boy walked towards the building.

Solution: Let AB be the building and CD be the initial position of the boy. Let C'D' be his final position. The boy's eyes are at point C. Let a horizontal line CE be drawn from his eyes to the building.

Height of the building, AB = 30 m. Height of the boy, CD = 1.5 m.

Height of the building above the boy's eye level, AE = AB - EB = AB - CD = 30 - 1.5 = 28.5 m.

Let C be the initial position and C' be the final position of the boy's eyes. Initial angle of elevation, $\angle ACE = 30^{\circ}$. Final angle of elevation, $\angle AC'E = 60^{\circ}$.

We need to find the distance walked, which is CC'. CC' = CE - C'E.

In the right-angled $\triangle ACE$: $\tan 30^{\circ} = \frac{AE}{CE}$ $\frac{1}{\sqrt{3}} = \frac{28.5}{CE}$ $CE = 28.5\sqrt{3} \text{ m}$

In the right-angled $\triangle AC'E$: $\tan 60^{\circ} = \frac{AE}{C'E}$ $\sqrt{3} = \frac{28.5}{C'E}$ $C'E = \frac{28.5}{\sqrt{3}} \text{ m}$

Distance walked = CC' = CE - C'E $= 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}}$ $= 28.5 \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)$ $= 28.5 \left( \frac{3 - 1}{\sqrt{3}} \right)$ $= 28.5 \times \frac{2}{\sqrt{3}} = \frac{57}{\sqrt{3}}$

To rationalize the denominator: $= \frac{57}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{57\sqrt{3}}{3} = 19\sqrt{3} \text{ m}$

Final Answer: The distance the boy walked towards the building is $19\sqrt{3}$ m.

To Find: The height of the transmission tower.

Solution: Let BC be the building and AB be the transmission tower on top of it. Let P be the point on the ground from where the observations are made. This forms two right-angled triangles, $\triangle PBC$ and $\triangle PAC$, with the right angle at C.

Height of the building, BC = 20 m. Let the height of the tower be AB = h. Total height from the ground to the top of the tower, AC = AB + BC = h + 20.

Angle of elevation to the bottom of the tower, $\angle BPC = 45^{\circ}$. Angle of elevation to the top of the tower, $\angle APC = 60^{\circ}$.

In the right-angled $\triangle PBC$: $\tan 45^{\circ} = \frac{BC}{PC}$ $1 = \frac{20}{PC}$ $PC = 20 \text{ m}$

Now, in the right-angled $\triangle PAC$: $\tan 60^{\circ} = \frac{AC}{PC}$ $\sqrt{3} = \frac{h + 20}{20}$ $20\sqrt{3} = h + 20$ $h = 20\sqrt{3} - 20$ $h = 20(\sqrt{3} - 1) \text{ m}$

Final Answer: The height of the tower is $20(\sqrt{3} - 1)$ m.

To Find: The height of the pedestal.

Solution: Let BC be the pedestal and AB be the statue on top of it. Let P be the point on the ground from where the observations are made. This forms two right-angled triangles, $\triangle PBC$ and $\triangle PAC$, with the right angle at C.

Height of the statue, AB = 1.6 m. Let the height of the pedestal be BC = h. Total height from the ground to the top of the statue, AC = AB + BC = 1.6 + h.

Angle of elevation to the top of the pedestal, $\angle BPC = 45^{\circ}$. Angle of elevation to the top of the statue, $\angle APC = 60^{\circ}$.

In the right-angled $\triangle PBC$: $\tan 45^{\circ} = \frac{BC}{PC}$ $1 = \frac{h}{PC}$ $PC = h$

Now, in the right-angled $\triangle PAC$: $\tan 60^{\circ} = \frac{AC}{PC}$ $\sqrt{3} = \frac{1.6 + h}{PC}$

Since PC = h, we can substitute it into the equation: $\sqrt{3} = \frac{1.6 + h}{h}$ $h\sqrt{3} = 1.6 + h$ $h\sqrt{3} - h = 1.6$ $h(\sqrt{3} - 1) = 1.6$ $h = \frac{1.6}{\sqrt{3} - 1}$

To rationalize the denominator, multiply the numerator and denominator by $(\sqrt{3} + 1)$: $h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$ $h = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$ $h = \frac{1.6(\sqrt{3} + 1)}{2}$ $h = 0.8(\sqrt{3} + 1) \text{ m}$

Final Answer: The height of the pedestal is $0.8(\sqrt{3} + 1)$ m.

To Find: The height of the building.

Solution: Let AB be the building and CD be the tower. They are on level ground, so BD is the horizontal distance between them.

Height of the tower, CD = 50 m. Let the height of the building be AB = h.

The angle of elevation of the top of the tower (C) from the foot of the building (B) is $\angle CBD = 60^{\circ}$. In the right-angled $\triangle CBD$: $\tan 60^{\circ} = \frac{CD}{BD}$ $\sqrt{3} = \frac{50}{BD}$ $BD = \frac{50}{\sqrt{3}} \text{ m}$

The angle of elevation of the top of the building (A) from the foot of the tower (D) is $\angle ADB = 30^{\circ}$. In the right-angled $\triangle ABD$: $\tan 30^{\circ} = \frac{AB}{BD}$ $\frac{1}{\sqrt{3}} = \frac{h}{BD}$ $h = \frac{BD}{\sqrt{3}}$

Now, substitute the value of BD from the first part into this equation: $h = \frac{\left(\frac{50}{\sqrt{3}}\right)}{\sqrt{3}}$ $h = \frac{50}{\sqrt{3} \times \sqrt{3}}$ $h = \frac{50}{3} \text{ m}$

Final Answer: The height of the building is $\frac{50}{3}$ m or $16 \frac{2}{3}$ m.

Solution: Let AB and CD be the two poles of equal height, h. Let the road be AC, so AC = 80 m. Let P be the point on the road between the poles.

Let the distance of the point P from pole AB be AP = x. Then the distance of the point P from pole CD will be PC = 80 - x.

The angle of elevation to the top of pole AB is $\angle APB = 60^{\circ}$. The angle of elevation to the top of pole CD is $\angle CPD = 30^{\circ}$.

In the right-angled $\triangle PAB$: $\tan 60^{\circ} = \frac{AB}{AP}$ $\sqrt{3} = \frac{h}{x}$ $h = x\sqrt{3} \quad \dots(1)$

In the right-angled $\triangle PCD$: $\tan 30^{\circ} = \frac{CD}{PC}$ $\frac{1}{\sqrt{3}} = \frac{h}{80 - x}$ $h = \frac{80 - x}{\sqrt{3}} \quad \dots(2)$

Since the heights are equal, we can equate (1) and (2): $x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$ $x\sqrt{3} \times \sqrt{3} = 80 - x$ $3x = 80 - x$ $4x = 80$ $x = 20 \text{ m}$

So, the distance of the point from the first pole is AP = 20 m. The distance of the point from the second pole is PC = 80 - 20 = 60 m.

Now, find the height h using equation (1): $h = x\sqrt{3} = 20\sqrt{3} \text{ m}$

Final Answer: The height of the poles is $20\sqrt{3}$ m. The distances of the point from the poles are 20 m and 60 m.

Solution: Let AB be the TV tower and BC be the width of the canal. C is the point on the other bank directly opposite the tower. Let D be the point 20 m away from C, such that C is between B and D. So, CD = 20 m.

Angle of elevation from C, $\angle ACB = 60^{\circ}$. Angle of elevation from D, $\angle ADB = 30^{\circ}$.

Let the height of the tower be AB = h and the width of the canal be BC = x.

In the right-angled $\triangle ABC$: $\tan 60^{\circ} = \frac{AB}{BC}$ $\sqrt{3} = \frac{h}{x}$ $h = x\sqrt{3} \quad \dots(1)$

In the right-angled $\triangle ABD$: The base is BD = BC + CD = x + 20. $\tan 30^{\circ} = \frac{AB}{BD}$ $\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$ $h = \frac{x + 20}{\sqrt{3}} \quad \dots(2)$

Equating (1) and (2) to find x: $x\sqrt{3} = \frac{x + 20}{\sqrt{3}}$ $x\sqrt{3} \times \sqrt{3} = x + 20$ $3x = x + 20$ $2x = 20$ $x = 10 \text{ m}$

So, the width of the canal is 10 m.

Now, substitute the value of x in equation (1) to find the height h: $h = x\sqrt{3} = 10\sqrt{3} \text{ m}$

Final Answer: The height of the tower is $10\sqrt{3}$ m and the width of the canal is 10 m.

To Find: The height of the cable tower.

Solution: Let AB be the building and CD be the cable tower. Let the height of the building be AB = 7 m. Draw a line AE parallel to the ground BD from the top of the building A.

Height of the building, AB = 7 m. So, ED = 7 m. Distance between the building and tower, BD = AE.

Angle of elevation of the top of the tower (C) from A is $\angle CAE = 60^{\circ}$. Angle of depression of the foot of the tower (D) from A is $\angle EAD = 45^{\circ}$. Since AE is parallel to BD, $\angle ADB = \angle EAD = 45^{\circ}$ (alternate interior angles).

In the right-angled $\triangle ABD$: $\tan 45^{\circ} = \frac{AB}{BD}$ $1 = \frac{7}{BD}$ $BD = 7 \text{ m}$

Since AE = BD, then AE = 7 m.

Now, in the right-angled $\triangle ACE$: $\tan 60^{\circ} = \frac{CE}{AE}$ $\sqrt{3} = \frac{CE}{7}$ $CE = 7\sqrt{3} \text{ m}$

The height of the tower is CD = CE + ED. $CD = 7\sqrt{3} + 7$ $CD = 7(\sqrt{3} + 1) \text{ m}$

Final Answer: The height of the tower is $7(\sqrt{3} + 1)$ m.

To Find: The distance between the two ships.

Solution: Let AB be the lighthouse with height AB = 75 m. Let C and D be the positions of the two ships such that D is behind C.

The angle of depression from the top of the lighthouse A to ship C is $45^{\circ}$. Therefore, the angle of elevation from C to A is $\angle ACB = 45^{\circ}$. The angle of depression from A to ship D is $30^{\circ}$. Therefore, the angle of elevation from D to A is $\angle ADB = 30^{\circ}$.

In the right-angled $\triangle ABC$: $\tan 45^{\circ} = \frac{AB}{BC}$ $1 = \frac{75}{BC}$ $BC = 75 \text{ m}$

In the right-angled $\triangle ABD$: $\tan 30^{\circ} = \frac{AB}{BD}$ $\frac{1}{\sqrt{3}} = \frac{75}{BD}$ $BD = 75\sqrt{3} \text{ m}$

The distance between the two ships is CD. CD = BD - BC $CD = 75\sqrt{3} - 75$ $CD = 75(\sqrt{3} - 1) \text{ m}$

Final Answer: The distance between the two ships is $75(\sqrt{3} - 1)$ m.

To Find: The distance travelled by the balloon.

Solution: Let the girl's position be at point G on the ground. Her eyes are at point E, so EG = 1.2 m. Let A be the initial position of the balloon and B be its final position. The balloon is moving in a horizontal line. Let C and D be points on the ground directly below A and B respectively. Draw a horizontal line EF parallel to the ground from the girl's eyes.

The height of the balloon from the ground is AC = BD = 88.2 m. The height of the balloon from the girl's eye level is AF = BF' = 88.2 - 1.2 = 87 m (where F and F' are points on EF below A and B).

Initial position: Angle of elevation $\angle AEF = 60^{\circ}$. Final position: Angle of elevation $\angle BEF' = 30^{\circ}$.

In the right-angled $\triangle AEF$: $\tan 60^{\circ} = \frac{AF}{EF}$ $\sqrt{3} = \frac{87}{EF}$ $EF = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3} \text{ m}$

In the right-angled $\triangle BEF'$: $\tan 30^{\circ} = \frac{BF'}{EF'}$ $\frac{1}{\sqrt{3}} = \frac{87}{EF'}$ $EF' = 87\sqrt{3} \text{ m}$

The distance travelled by the balloon is AB, which is equal to the horizontal distance FF'. Distance = FF' = EF' - EF $= 87\sqrt{3} - 29\sqrt{3}$ $= (87 - 29)\sqrt{3}$ $= 58\sqrt{3} \text{ m}$

Final Answer: The distance travelled by the balloon during the interval is $58\sqrt{3}$ m.

To Find: Time taken by the car to reach the foot of the tower from the second point.

Solution: Let AB be the tower of height h. Let C be the initial position of the car and D be its position after 6 seconds.

The angle of depression from A to C is $30^{\circ}$, so the angle of elevation $\angle ACB = 30^{\circ}$. The angle of depression from A to D is $60^{\circ}$, so the angle of elevation $\angle ADB = 60^{\circ}$.

In the right-angled $\triangle ABD$: $\tan 60^{\circ} = \frac{AB}{BD}$ $\sqrt{3} = \frac{h}{BD}$ $BD = \frac{h}{\sqrt{3}} \quad \dots(1)$

In the right-angled $\triangle ABC$: $\tan 30^{\circ} = \frac{AB}{BC}$ $\frac{1}{\sqrt{3}} = \frac{h}{BC}$ $BC = h\sqrt{3} \quad \dots(2)$

The distance covered by the car in 6 seconds is CD. CD = BC - BD $CD = h\sqrt{3} - \frac{h}{\sqrt{3}}$ $CD = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = h \left( \frac{3-1}{\sqrt{3}} \right) = \frac{2h}{\sqrt{3}}$

Let the uniform speed of the car be s. Speed, $s = \frac{\text{Distance}}{\text{Time}} = \frac{CD}{6} = \frac{2h/\sqrt{3}}{6} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}}$

Now, we need to find the time taken to travel the remaining distance BD. Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{BD}{s}$

Substitute the expressions for BD and s: $\text{Time} = \frac{h/\sqrt{3}}{h/(3\sqrt{3})}$ $\text{Time} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h}$ $\text{Time} = 3 \text{ seconds}$

Final Answer: The time taken by the car to reach the foot of the tower from this point is 3 seconds.