CBSE Class 10 Mathematics Trigonometry Assignment Set 18

Question. At a point on the ground, the angles of elevation of a 10 m tall building and a helicopter hovering some distance over the top of the building are \( 45^\circ \) and \( 60^\circ \) respectively. Find the height of the helicopter above the ground.
Answer: Let \( AB \) denote the building of height \( 10\text{ m} \), where \( A \) is the roof and \( B \) is the base on the ground. Let \( H \) be the helicopter hovering directly over the building. Let \( P \) be the observer's point on the ground.
According to the problem:
The height of the building, \( AB = 10\text{ m} \).
The angle of elevation of the top of the building \( A \) from point \( P \) is \( 45^\circ \). Therefore, \( \angle APB = 45^\circ \).
The angle of elevation of the helicopter \( H \) from point \( P \) is \( 60^\circ \). Therefore, \( \angle HPB = 60^\circ \).

In the right-angled triangle \( \triangle ABP \):
\( \tan(45^\circ) = \frac{AB}{PB} \)

\( \implies 1 = \frac{10}{PB} \)

\( \implies PB = 10\text{ m} \)

In the right-angled triangle \( \triangle HBP \):
\( \tan(60^\circ) = \frac{HB}{PB} \)

\( \implies \sqrt{3} = \frac{HB}{10} \)

\( \implies HB = 10\sqrt{3}\text{ m} \)

Using \( \sqrt{3} \approx 1.732 \):
\( HB = 10 \times 1.732 = 17.32\text{ m} \)

Therefore, the height of the helicopter above the ground is \( 10\sqrt{3}\text{ m} \) (or approximately \( 17.32\text{ m} \)).
In simple words: First, we use the 45-degree angle to find the observer's horizontal distance from the building, which is equal to the building's height. Then, we use the 60-degree angle to compute the helicopter's total height from the ground.
Exam Tip: Whenever you work with a \( 45^\circ \) angle of elevation, the base and perpendicular lengths are equal. Always solve this triangle first to find the horizontal distance quickly.

Question. A balloon at a height of 50 m is moving with the wind in a horizontal line. The angle of elevation of the balloon from the eye of the observer at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during this interval.
Answer: Let the observer be located at point \( O \). Let the balloon travel horizontally at a constant height of \( 50\text{ m} \) above the ground line.
Initially, the balloon is at position \( A \), and its horizontal line projection is at point \( P \). Thus, \( AP = 50\text{ m} \). The angle of elevation is \( 60^\circ \), which means \( \angle AOP = 60^\circ \).
After moving with the wind, the balloon reaches position \( B \), and its projection on the horizontal line is at point \( Q \). Thus, \( BQ = 50\text{ m} \). The new angle of elevation is \( 30^\circ \), so \( \angle BOQ = 30^\circ \).

In the right-angled triangle \( \triangle APO \):
\( \tan(60^\circ) = \frac{AP}{OP} \)

\( \implies \sqrt{3} = \frac{50}{OP} \)

\( \implies OP = \frac{50}{\sqrt{3}}\text{ m} \)

In the right-angled triangle \( \triangle BQO \):
\( \tan(30^\circ) = \frac{BQ}{OQ} \)

\( \implies \frac{1}{\sqrt{3}} = \frac{50}{OQ} \)

\( \implies OQ = 50\sqrt{3}\text{ m} \)

The horizontal distance covered by the balloon is \( PQ \):
\( PQ = OQ - OP \)

\( \implies PQ = 50\sqrt{3} - \frac{50}{\sqrt{3}} \)

\( \implies PQ = \frac{150 - 50}{\sqrt{3}} = \frac{100}{\sqrt{3}}\text{ m} \)

Rationalizing the denominator:
\( PQ = \frac{100\sqrt{3}}{3}\text{ m} \)

Using the approximation \( \sqrt{3} \approx 1.732 \):
\( PQ = \frac{100 \times 1.732}{3} \approx 57.74\text{ m} \)

The distance covered by the balloon during this interval is \( \frac{100\sqrt{3}}{3}\text{ m} \) (or approximately \( 57.74\text{ m} \)).
In simple words: We find the horizontal distance to the balloon at both angles of elevation. Subtracting the smaller horizontal distance from the larger one gives us the total horizontal distance the balloon traveled.

Exam Tip: Remember that as an object moves farther away horizontally at a fixed altitude, its angle of elevation decreases. Always subtract the smaller distance from the larger distance.

Question. A straight road leads to the foot of the tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Answer: Let \( AB \) represent the tower of height \( h \), with \( A \) being its top and \( B \) being its foot. Let \( C \) and \( D \) be the initial and final positions of the car on the road, respectively, after an interval of \( 6\text{ seconds} \).
The angles of depression of the car from point \( A \) at positions \( C \) and \( D \) are \( 30^\circ \) and \( 60^\circ \). By alternate interior angles, we have:
\( \angle ACB = 30^\circ \) and \( \angle ADB = 60^\circ \).

In the right-angled triangle \( \triangle ABD \):
\( \tan(60^\circ) = \frac{AB}{BD} \)

\( \implies \sqrt{3} = \frac{h}{BD} \)

\( \implies BD = \frac{h}{\sqrt{3}} \)

In the right-angled triangle \( \triangle ABC \):
\( \tan(30^\circ) = \frac{AB}{BC} \)

\( \implies \frac{1}{\sqrt{3}} = \frac{h}{BC} \)

\( \implies BC = h\sqrt{3} \)

The distance covered by the car in \( 6\text{ seconds} \) is \( CD \):
\( CD = BC - BD \)

\( \implies CD = h\sqrt{3} - \frac{h}{\sqrt{3}} \)

\( \implies CD = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \)

Since the car is moving at a uniform speed \( v \):
\( v = \frac{\text{Distance}}{\text{Time}} = \frac{CD}{6} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}}\text{ units per second} \)

The time taken to cover the remaining distance \( BD \) is:
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{BD}{v} \)

\( \implies \text{Time} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} = 3\text{ seconds} \)

The car takes \( 3\text{ seconds} \) to reach the foot of the tower from point \( D \).
In simple words: The distance from the second observation point to the tower is exactly half the distance covered in the first 6 seconds. Since the speed is uniform, it takes exactly half the time, which is 3 seconds.

Exam Tip: When an object moves closer to a tower, the angle of depression increases. Make sure to put the larger angle (\( 60^\circ \)) closer to the base of the tower.

Question. From the top of a 10m building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Answer: Let \( AB \) represent the building of height \( 10\text{ m} \), where \( A \) is its top and \( B \) is its bottom. Let \( CD \) be the cable tower of height \( H \), where \( C \) is its top and \( D \) is its foot.
Draw a horizontal line from point \( A \) meeting the cable tower perpendicularly at point \( E \). This forms a rectangle \( ABDE \), meaning:
\( ED = AB = 10\text{ m} \)
\( AE = BD \)

According to the given information:
The angle of elevation of the top of the tower \( C \) from point \( A \) is \( 60^\circ \), so \( \angle CAE = 60^\circ \).
The angle of depression of the base of the tower \( D \) from point \( A \) is \( 45^\circ \). By alternate angles, \( \angle ADB = 45^\circ \) (or \( \angle EAD = 45^\circ \)).

In the right-angled triangle \( \triangle AED \):
\( \tan(45^\circ) = \frac{ED}{AE} \)

\( \implies 1 = \frac{10}{AE} \)

\( \implies AE = 10\text{ m} \)

Since \( AE = BD \), the distance between the building and the cable tower is \( 10\text{ m} \).

In the right-angled triangle \( \triangle CAE \):
\( \tan(60^\circ) = \frac{CE}{AE} \)

\( \implies \sqrt{3} = \frac{CE}{10} \)

\( \implies CE = 10\sqrt{3}\text{ m} \)

The total height of the cable tower is \( CD \):
\( CD = CE + ED \)

\( \implies CD = 10\sqrt{3} + 10 = 10(\sqrt{3} + 1)\text{ m} \)

Using \( \sqrt{3} \approx 1.732 \):
\( CD \approx 10(1.732 + 1) = 10 \times 2.732 = 27.32\text{ m} \)

The height of the cable tower is \( 10(\sqrt{3} + 1)\text{ m} \) (or approximately \( 27.32\text{ m} \)).
In simple words: Using the 45-degree angle of depression, we find that the distance between the building and the tower is equal to the building's height (10 meters). We then use this distance with the 60-degree angle to find the upper height of the tower and combine both sections.

Exam Tip: When working with combined elevation and depression angles, draw a horizontal baseline from the observer's eye to break the height of the opposite object into two clear segments.

Question. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height h. At a point on the plane, the angle of elevation of the top and the bottom of the flag-staff is \( \beta \) and \( \alpha \) respectively. Prove that the height of the tower is \( \frac { h\tan \alpha }{ \tan \beta -\tan \alpha } \)
Answer: Let \( BC \) represent the vertical tower of height \( y \), and let \( CD \) represent the vertical flag-staff of height \( h \) mounted on top of it.
Let \( P \) be an observation point on the horizontal plane at a distance \( x \) from the base of the tower \( B \).
The angle of elevation of the bottom of the flag-staff \( C \) from point \( P \) is \( \alpha \). Therefore, \( \angle CPB = \alpha \).
The angle of elevation of the top of the flag-staff \( D \) from point \( P \) is \( \beta \). Therefore, \( \angle DPB = \beta \).

In the right-angled triangle \( \triangle CBP \):
\( \tan \alpha = \frac{BC}{PB} = \frac{y}{x} \)

\( \implies x = \frac{y}{\tan \alpha} \) --- (1)

In the right-angled triangle \( \triangle DBP \):
\( \tan \beta = \frac{BD}{PB} = \frac{y + h}{x} \)

\( \implies x = \frac{y + h}{\tan \beta} \) --- (2)

Equating equations (1) and (2) as they both express the horizontal distance \( x \):
\( \frac{y}{\tan \alpha} = \frac{y + h}{\tan \beta} \)

\( \implies y \tan \beta = (y + h) \tan \alpha \)

\( \implies y \tan \beta = y \tan \alpha + h \tan \alpha \)

\( \implies y \tan \beta - y \tan \alpha = h \tan \alpha \)

\( \implies y(\tan \beta - \tan \alpha) = h \tan \alpha \)

\( \implies y = \frac{h \tan \alpha}{\tan \beta - \tan \alpha} \)

The height of the tower is proven to be \( \frac{h \tan \alpha}{\tan \beta - \tan \alpha} \).
In simple words: We form expressions for the horizontal distance using both triangles. Since this horizontal distance is identical in both cases, we equate them and rearrange terms to solve for the tower's height.

Exam Tip: In algebraic proof questions, keep all variables general and do not insert estimated numbers. Group the target variable terms on one side of the equation systematically.

Question. From the top and bottom of a building of height h, the angles of elevation of a tower are \( \theta \) and \( \phi \) respectively. Show that the height of the tower is \( \frac { h\tan \phi }{ \tan \phi -\tan \theta } \)
Answer: Let \( AB \) represent the building of height \( h \), and let \( CD \) be the tower of height \( H \). Let the horizontal distance between the building and the tower be \( d \).
Draw a horizontal line from point \( A \) (top of the building) to meet the tower perpendicularly at point \( E \). This horizontal line provides:
\( AE = BD = d \)
\( ED = AB = h \)
\( CE = CD - ED = H - h \)

From the bottom of the building \( B \), the angle of elevation of the top of the tower \( C \) is \( \phi \), so \( \angle CBD = \phi \).
From the top of the building \( A \), the angle of elevation of the top of the tower \( C \) is \( \theta \), so \( \angle CAE = \theta \).

In the right-angled triangle \( \triangle CBD \):
\( \tan \phi = \frac{CD}{BD} = \frac{H}{d} \)

\( \implies d = \frac{H}{\tan \phi} \) --- (1)

In the right-angled triangle \( \triangle CAE \):
\( \tan \theta = \frac{CE}{AE} = \frac{H - h}{d} \)

\( \implies d = \frac{H - h}{\tan \theta} \) --- (2)

Equating the two expressions for \( d \) from equations (1) and (2):
\( \frac{H}{\tan \phi} = \frac{H - h}{\tan \theta} \)

\( \implies H \tan \theta = (H - h) \tan \phi \)

\( \implies H \tan \theta = H \tan \phi - h \tan \phi \)

\( \implies h \tan \phi = H \tan \phi - H \tan \theta \)

\( \implies h \tan \phi = H(\tan \phi - \tan \theta) \)

\( \implies H = \frac{h \tan \phi}{\tan \phi - \tan \theta} \)

The height of the tower is proven to be \( \frac{h \tan \phi}{\tan \phi - \tan \theta} \).
In simple words: We find the horizontal distance between the building and the tower from both right-angled triangles. Setting these two expressions equal allows us to solve for the tower's full height.

Exam Tip: Be careful with alternate interior angles when dealing with multiple observation points. Draw horizontal lines to trace angles accurately.

Question. From the top of a light - house, the angle of depression of two ships on the opposite sides of it are observed to be \( \alpha \) and \( \beta \). of the height of the light house be h metres and the live joining the ships passes through the foot of the light house. show that the distance between the ships is \( \frac { h(\tan \alpha +\tan \beta ) }{ \tan \alpha .\tan \beta } \)
Answer: Let \( AB \) represent the lighthouse of height \( h \), where \( A \) is its top and \( B \) is its base on the ground. Let \( C \) and \( D \) denote the positions of the two ships on opposite sides of the lighthouse such that \( C, B, D \) lie on a straight line.
The angles of depression of the ships \( C \) and \( D \) from the top of the lighthouse \( A \) are \( \alpha \) and \( \beta \) respectively. By alternate interior angles:
\( \angle ACB = \alpha \) and \( \angle ADB = \beta \).

In the right-angled triangle \( \triangle ABC \):
\( \tan \alpha = \frac{AB}{BC} = \frac{h}{BC} \)

\( \implies BC = \frac{h}{\tan \alpha} \)

In the right-angled triangle \( \triangle ABD \):
\( \tan \beta = \frac{AB}{BD} = \frac{h}{BD} \)

\( \implies BD = \frac{h}{\tan \beta} \)

The total distance between the two ships is \( CD \):
\( CD = BC + BD \)

\( \implies CD = \frac{h}{\tan \alpha} + \frac{h}{\tan \beta} \)

\( \implies CD = h \left( \frac{1}{\tan \alpha} + \frac{1}{\tan \beta} \right) \)

\( \implies CD = h \left( \frac{\tan \alpha + \tan \beta}{\tan \alpha \cdot \tan \beta} \right) \)

The distance between the ships is proven to be \( \frac{h(\tan \alpha + \tan \beta)}{\tan \alpha \cdot \tan \beta} \).
In simple words: We calculate each ship's horizontal distance from the base of the lighthouse separately. Adding these two distances together and finding a common denominator yields the required proof.

Exam Tip: If objects are on opposite sides of a central line, you add their horizontal distances. If they are on the same side, you subtract them. Pay close attention to this distinction in exams.

Question. The angles of elevation of the top of a tower from two points at a distance of 'a' and 'b' from the base of the tower and in the same straight line with it are complementary. Prove that the height of tower is \( \sqrt { ab } \).
Answer: Let \( AB \) represent the tower of height \( h \), with \( A \) being the top and \( B \) being its base. Let \( C \) and \( D \) be two points on the ground in a straight line with \( B \).
The distance of point \( C \) from \( B \) is \( BC = a \).
The distance of point \( D \) from \( B \) is \( BD = b \).
The angles of elevation of point \( A \) from \( C \) and \( D \) are given as complementary.
If the angle of elevation at \( C \) is \( \theta \) (so \( \angle ACB = \theta \)), then the angle of elevation at \( D \) must be \( 90^\circ - \theta \) (so \( \angle ADB = 90^\circ - \theta \)).

In the right-angled triangle \( \triangle ABC \):
\( \tan \theta = \frac{AB}{BC} = \frac{h}{a} \) --- (1)

In the right-angled triangle \( \triangle ABD \):
\( \tan(90^\circ - \theta) = \frac{AB}{BD} = \frac{h}{b} \)
Since \( \tan(90^\circ - \theta) = \cot \theta \):
\( \cot \theta = \frac{h}{b} \) --- (2)

Multiplying equation (1) and equation (2):
\( \tan \theta \cdot \cot \theta = \frac{h}{a} \cdot \frac{h}{b} \)
Since \( \tan \theta \cdot \cot \theta = 1 \):
\( 1 = \frac{h^2}{ab} \)

\( \implies h^2 = ab \)

\( \implies h = \sqrt{ab} \) (taking the positive root as height cannot be negative)

The height of the tower is proven to be \( \sqrt{ab} \).

In simple words: When two angles are complementary, their tangents multiply to 1. Writing both equations and multiplying them together simplifies the expressions and yields the square of the height.

Exam Tip: Keep in mind that \( \tan(90^\circ - \theta) = \cot \theta \). Multiplying \( \tan \theta \) by \( \cot \theta \) will always eliminate the trigonometric ratio, which is a powerful step in proof problems.