CBSE Class 10 Mathematics Trigonometry Assignment Set 17

Question. The angle of elevation of the top of a tower at a point, at a horizontal distance 100m from its foot is 30°. Find the height of the tower.
Answer: Let the height of the tower be denoted as \( h \) meters. The horizontal distance from the foot of the tower is given as 100 m. The angle of elevation to the top of the tower is \( 30^\circ \). In the right-angled triangle formed: \( \tan 30^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{100} \) Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we have: \( \frac{1}{\sqrt{3}} = \frac{h}{100} \)
\( \implies h = \frac{100}{\sqrt{3}}\text{ m} \) Rationalizing the denominator gives: \( h = \frac{100\sqrt{3}}{3}\text{ m} \) Using the approximation \( \sqrt{3} \approx 1.732 \): \( h \approx \frac{100 \times 1.732}{3} = \frac{173.2}{3} \approx 57.73\text{ m} \) Consequently, the height of the tower is \( \frac{100}{\sqrt{3}}\text{ m} \) (which is approximately \( 57.73\text{ m} \)).
In simple words: Imagine a right-angled triangle where the ground distance is 100 meters and the angle is 30 degrees. Using the tangent function, we find that the height of the tower is \( \frac{100}{\sqrt{3}} \) meters, or about 57.73 meters.

Exam Tip: Always sketch a neat, labeled right-angled triangle first. Clearly representing the unknown height \( h \), the given distance, and the angle of elevation helps secure easy step-marks.

Question. The shadow of a tower, standing on a level ground, is found to be 45m longer when Sun's altitude is 30° than when it was 60°. Find the height of the tower.
Answer: Let \( h \) represent the height of the tower, and \( x \) be the length of its shadow when the angle of elevation is \( 60^\circ \). When the elevation of the Sun decreases to \( 30^\circ \), the shadow length becomes \( x + 45 \) meters. In the right-angled triangle corresponding to the \( 60^\circ \) angle: \( \tan 60^\circ = \frac{h}{x} \) Since \( \tan 60^\circ = \sqrt{3} \): \( \sqrt{3} = \frac{h}{x} \)
\( \implies x = \frac{h}{\sqrt{3}} \) - (Equation 1) In the right-angled triangle corresponding to the \( 30^\circ \) angle: \( \tan 30^\circ = \frac{h}{x + 45} \) Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \): \( \frac{1}{\sqrt{3}} = \frac{h}{x + 45} \)
\( \implies x + 45 = h\sqrt{3} \) - (Equation 2) Substituting Equation 1 into Equation 2: \( \frac{h}{\sqrt{3}} + 45 = h\sqrt{3} \) Multiplying the entire equation by \( \sqrt{3} \) to eliminate the fraction: \( h + 45\sqrt{3} = 3h \)
\( \implies 2h = 45\sqrt{3} \)
\( \implies h = 22.5\sqrt{3}\text{ m} \) Taking \( \sqrt{3} \approx 1.732 \): \( h \approx 22.5 \times 1.732 = 38.97\text{ m} \) Therefore, the height of the tower is \( 22.5\sqrt{3}\text{ m} \) (or approximately \( 38.97\text{ m} \)).
In simple words: As the angle of the Sun drops from 60 degrees to 30 degrees, the shadow lengthens by 45 meters. Using trigonometric equations for both conditions, we find that the tower is \( 22.5\sqrt{3} \) meters tall, which is about 38.97 meters.

Exam Tip: Be sure not to mix up the angles — a larger angle of elevation (\( 60^\circ \)) corresponds to a shorter shadow closer to the base, while a smaller angle (\( 30^\circ \)) produces a longer shadow further away.

Question. From the top of a hill 200m high, the angles of depression of the top and the bottom of a tower are observed to be 30° and 60° respectively. Find the height of the tower and horizontal distance between them.
Answer: Let \( AB = 200\text{ m} \) represent the hill, with \( A \) being its peak. Let \( CD = h \) represent the tower, with \( C \) being its top. Let the horizontal distance between the hill and the tower be \( BD = d \). Draw a horizontal reference line from \( C \) meeting \( AB \) perpendicularly at point \( E \). Thus, \( EC = BD = d \) and \( EB = CD = h \). This leaves the segment \( AE = AB - EB = 200 - h \). According to the given observations of depression: The angle of depression to the bottom of the tower \( D \) is \( 60^\circ \), which means \( \angle ADB = 60^\circ \). In the right-angled triangle \( ABD \): \( \tan 60^\circ = \frac{AB}{BD} \) \( \sqrt{3} = \frac{200}{d} \)
\( \implies d = \frac{200}{\sqrt{3}}\text{ m} \) Rationalizing the denominator: \( d = \frac{200\sqrt{3}}{3}\text{ m} \) Taking \( \sqrt{3} \approx 1.732 \): \( d \approx \frac{200 \times 1.732}{3} = \frac{346.4}{3} \approx 115.47\text{ m} \) The angle of depression to the top of the tower \( C \) is \( 30^\circ \), which means \( \angle ACE = 30^\circ \). In the right-angled triangle \( AEC \): \( \tan 30^\circ = \frac{AE}{EC} \) \( \frac{1}{\sqrt{3}} = \frac{200 - h}{d} \) Substituting our expression for \( d \): \( \frac{1}{\sqrt{3}} = \frac{200 - h}{200/\sqrt{3}} \) \( \frac{200}{3} = 200 - h \)
\( \implies h = 200 - \frac{200}{3} = \frac{400}{3}\text{ m} \approx 133.33\text{ m} \) Therefore: The height of the tower is \( \frac{400}{3}\text{ m} \) (or approximately \( 133.33\text{ m} \)). The horizontal distance between them is \( \frac{200\sqrt{3}}{3}\text{ m} \) (or approximately \( 115.47\text{ m} \)).
In simple words: Looking down from a 200-meter hill, the angle of depression to the bottom of the tower helps us calculate that the ground distance is 115.47 meters. Using the angle to the top of the tower, we find that the tower itself is 133.33 meters tall.

Exam Tip: Remember that the angle of depression is measured from an imaginary horizontal line of sight looking downward. Draw this horizontal line at the top to clearly indicate how the angles of depression relate to the angles inside the triangles.

Question. Two poles of equal height stand on either side of a road 150m wide. At a point in between the poles, the elevations of the tops of the poles are 60° and 30°. Find their height and position of the point.
Answer: Let \( AB \) and \( CD \) be the two poles of equal height \( h \) meters situated on both sides of a road of width \( BD = 150\text{ m} \). Let \( P \) be the point on the road such that the distance \( BP = x \) meters and \( PD = 150 - x \) meters. The angles of elevation from point \( P \) to the tops of the poles \( AB \) and \( CD \) are \( 60^\circ \) and \( 30^\circ \) respectively. In the right-angled triangle \( ABP \): \( \tan 60^\circ = \frac{AB}{BP} \) \( \sqrt{3} = \frac{h}{x} \)
\( \implies h = x\sqrt{3} \) - (Equation 1) In the right-angled triangle \( CDP \): \( \tan 30^\circ = \frac{CD}{PD} \) \( \frac{1}{\sqrt{3}} = \frac{h}{150 - x} \)
\( \implies 150 - x = h\sqrt{3} \) - (Equation 2) Substituting the value of \( h \) from Equation 1 into Equation 2: \( 150 - x = (x\sqrt{3})\sqrt{3} \) \( 150 - x = 3x \) \( 4x = 150 \)
\( \implies x = 37.5\text{ m} \) Thus, the point is located \( 37.5\text{ m} \) away from the first pole (which has an elevation of \( 60^\circ \)) and \( 112.5\text{ m} \) away from the second pole. To find the height of the poles, substitute \( x = 37.5 \) back into Equation 1: \( h = 37.5\sqrt{3}\text{ m} \) Taking \( \sqrt{3} \approx 1.732 \): \( h \approx 37.5 \times 1.732 = 64.95\text{ m} \) Therefore, the height of each pole is \( 37.5\sqrt{3}\text{ m} \) (or approximately \( 64.95\text{ m} \)).
In simple words: The road is 150 meters wide. The point is closer to one pole (37.5 meters) than the other (112.5 meters). By utilizing trigonometry, we find that both of the equal-sized poles are about 64.95 meters high.

Exam Tip: Since both poles are of equal height, the point on the ground must be closer to the pole with the larger angle of elevation (\( 60^\circ \)). Always define the position of the point clearly with respect to both poles to avoid losing minor marks.

Question. The upper part of a tree broken over by the wind makes an angle of 60° with the ground and the distance from the root to the point where the top touches the ground is 20 m. Find the original height of the tree.
Answer: Let \( AB = h_1 \) represent the standing vertical part of the tree. Let \( AC = h_2 \) represent the broken section of the tree that has fallen, acting as the hypotenuse. The broken top of the tree touches the ground at point \( C \). The distance from the foot of the tree \( B \) to point \( C \) is \( BC = 20\text{ m} \). The angle between the fallen top and the ground is \( \angle ACB = 60^\circ \). In the right-angled triangle \( ABC \): To find the standing vertical height \( h_1 \): \( \tan 60^\circ = \frac{AB}{BC} \) \( \sqrt{3} = \frac{h_1}{20} \)
\( \implies h_1 = 20\sqrt{3}\text{ m} \) To find the length of the broken hypotenuse part \( h_2 \): \( \cos 60^\circ = \frac{BC}{AC} \) \( \frac{1}{2} = \frac{20}{h_2} \)
\( \implies h_2 = 40\text{ m} \) The original total height of the tree is the sum of both the standing and broken parts: \( H = h_1 + h_2 = 20\sqrt{3} + 40 = 20(2 + \sqrt{3})\text{ m} \) Using the value \( \sqrt{3} \approx 1.732 \): \( H \approx 20(1.732) + 40 = 34.64 + 40 = 74.64\text{ m} \) Therefore, the original height of the tree was \( 20(2 + \sqrt{3})\text{ m} \) (or approximately \( 74.64\text{ m} \)).
In simple words: A tree breaks in two. The portion still standing is 34.64 meters tall, and the broken top section that fell to the ground is 40 meters long. Adding these two together gives the original tree height of 74.64 meters.

Exam Tip: A very common error is calculating only the standing vertical height \( h_1 \). Always read carefully and sum both the vertical side and the hypotenuse to find the "original" height of the tree.

Question. The angles of depression of the top and bottom of a 75m tall building from the top of a tower are 45° and 60° resp. Find height of tower.
Answer: Let \( AB = H \) be the height of the tower, where \( A \) represents the top. Let \( CD = 75\text{ m} \) be the height of the building. Let the horizontal distance between them be \( BD = x \) meters. Draw a horizontal line from the top of the building \( C \) meeting \( AB \) at point \( E \). This gives \( EC = BD = x \) and \( EB = CD = 75\text{ m} \). The remaining upper part of the tower is \( AE = AB - EB = H - 75 \). Since the angle of depression of the bottom of the building \( D \) from \( A \) is \( 60^\circ \): In the right-angled triangle \( ABD \): \( \tan 60^\circ = \frac{AB}{BD} \) \( \sqrt{3} = \frac{H}{x} \)
\( \implies x = \frac{H}{\sqrt{3}} \) - (Equation 1) Since the angle of depression of the top of the building \( C \) from \( A \) is \( 45^\circ \): In the right-angled triangle \( AEC \): \( \tan 45^\circ = \frac{AE}{EC} \) \( 1 = \frac{H - 75}{x} \)
\( \implies x = H - 75 \) - (Equation 2) Equating the two expressions for \( x \): \( \frac{H}{\sqrt{3}} = H - 75 \) \( H = H\sqrt{3} - 75\sqrt{3} \) \( H(\sqrt{3} - 1) = 75\sqrt{3} \) \( H = \frac{75\sqrt{3}}{\sqrt{3} - 1}\text{ m} \) Multiplying both the numerator and the denominator by \( (\sqrt{3} + 1) \) to rationalize: \( H = \frac{75\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \) \( H = \frac{75(3 + \sqrt{3})}{2} = 37.5(3 + \sqrt{3})\text{ m} \) Using the approximation \( \sqrt{3} \approx 1.732 \): \( H \approx 37.5(3 + 1.732) = 37.5(4.732) = 177.45\text{ m} \) Therefore, the height of the tower is \( 37.5(3 + \sqrt{3})\text{ m} \) (which is about \( 177.45\text{ m} \)).
In simple words: By comparing the 45-degree and 60-degree angles of depression to the building's top and bottom, we establish two linked equations. Solving them reveals the height of the tower to be 177.45 meters.

Exam Tip: Rationalizing the denominator is essential. Do not leave fractions containing surds like \( \frac{1}{\sqrt{3}-1} \) in your final answer, as examiners look for simplified rationalized forms.

Question. An observer on the top of a lighthouse, 200 m tall, observes the angle of depression of two ships approaching the lighthouse to be 30° and 45° respectively. Find the distance between the two ships if they are (i) on the opposite side of the lighthouse (ii) on the same side of the lighthouse.
Answer: Let the lighthouse be \( AB = 200\text{ m} \), where \( A \) is the top. Let the positions of the two ships be \( C \) and \( D \). The angles of depression to the ships are \( 30^\circ \) and \( 45^\circ \) respectively, meaning their angles of elevation to the top are \( 30^\circ \) and \( 45^\circ \). Let the distance of the closer ship (elevation angle \( 45^\circ \)) from the base of the lighthouse be \( x \) meters: In right-angled triangle \( ABP \): \( \tan 45^\circ = \frac{AB}{x} \) \( 1 = \frac{200}{x} \implies x = 200\text{ m} \) Let the distance of the farther ship (elevation angle \( 30^\circ \)) from the base of the lighthouse be \( y \) meters: In right-angled triangle \( ABQ \): \( \tan 30^\circ = \frac{AB}{y} \) \( \frac{1}{\sqrt{3}} = \frac{200}{y} \implies y = 200\sqrt{3}\text{ m} \) Now we analyze the two required cases:
(i) **When they are on opposite sides of the lighthouse:** The distance between the two ships is the sum of their distances from the base: \( d_1 = y + x = 200\sqrt{3} + 200 = 200(\sqrt{3} + 1)\text{ m} \) Substituting \( \sqrt{3} \approx 1.732 \): \( d_1 \approx 200(1.732 + 1) = 200(2.732) = 546.4\text{ m} \)
(ii) **When they are on the same side of the lighthouse:** The distance between the two ships is the difference between their distances from the base: \( d_2 = y - x = 200\sqrt{3} - 200 = 200(\sqrt{3} - 1)\text{ m} \) Substituting \( \sqrt{3} \approx 1.732 \): \( d_2 \approx 200(1.732 - 1) = 200(0.732) = 146.4\text{ m} \) Consequently, the distance between the ships is \( 546.4\text{ m} \) if they are on opposite sides, and \( 146.4\text{ m} \) if they are on the same side.
In simple words: The first ship is 200 meters away, and the second ship is about 346.4 meters away. If they are on opposite sides, we add the distances to get 546.4 meters. If they are on the same side, we subtract them to get 146.4 meters.

Exam Tip: Be sure to explicitly write down and solve both scenarios (i) and (ii). Misreading "opposite side" and "same side" is a very frequent source of dropped marks in multi-part questions.

Question. The length of a string between a kite and a point on the ground is 150 m. If the string makes an angle θ with the level ground such that tan θ = 8/15 how high is the kite? Assume there is no slack in the string.
Answer: Let \( K \) represent the position of the kite and \( P \) be the point on the ground where the string is tied. The length of the string is the hypotenuse \( PK = 150\text{ m} \). Let the vertical height of the kite be \( h = KH \), where \( H \) is the projection of the kite on the level ground. The string's angle with the ground is \( \theta \). We are given: \( \tan\theta = \frac{8}{15} \) Since \( \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} \), we can find the hypotenuse of this reference ratio using Pythagoras' theorem: \( \text{Hypotenuse} = \sqrt{(\text{Opposite})^2 + (\text{Adjacent})^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \) This allows us to find the sine of the angle: \( \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8}{17} \) In the right-angled triangle \( KHP \): \( \sin\theta = \frac{KH}{PK} \) \( \frac{8}{17} = \frac{h}{150} \)
\( \implies h = \frac{150 \times 8}{17} = \frac{1200}{17}\text{ m} \approx 70.59\text{ m} \) Therefore, the height of the kite is \( \frac{1200}{17}\text{ m} \) (or approximately \( 70.59\text{ m} \)).
In simple words: The string is 150 meters long. Based on the given ratio \( \tan\theta = 8/15 \), we determine that the vertical ratio \( \sin\theta \) is \( 8/17 \). Multiplying this by the string's length gives a flying height of 70.59 meters.

Exam Tip: Do not waste time computing the angle \( \theta \) in degrees. It is far quicker and more precise to convert \( \tan\theta \) to \( \sin\theta \) algebraically using Pythagorean triplets.

Question. The angle of elevation of the top and bottom of a flag-staff fixed on a cliff are 60° and 45° to a person standing on the other end of a road 75 m wide. Find the height of the flag-staff.
Answer: Let \( BC = h \) represent the height of the cliff. Let \( AB = y \) represent the height of the flag-staff mounted on top of the cliff. The observer is standing at point \( D \), which is at a distance of \( CD = 75\text{ m} \) (the width of the road) from the foot of the cliff \( C \). The angles of elevation of the bottom \( B \) and top \( A \) of the flag-staff are \( 45^\circ \) and \( 60^\circ \) respectively. In the right-angled triangle \( BCD \): \( \tan 45^\circ = \frac{BC}{CD} \) \( 1 = \frac{h}{75} \implies h = 75\text{ m} \) In the right-angled triangle \( ACD \): \( \tan 60^\circ = \frac{AC}{CD} \) \( \sqrt{3} = \frac{h + y}{75} \) Substituting \( h = 75 \): \( 75\sqrt{3} = 75 + y \)
\( \implies y = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)\text{ m} \) Taking \( \sqrt{3} \approx 1.732 \): \( y \approx 75(1.732 - 1) = 75 \times 0.732 = 54.9\text{ m} \) Thus, the height of the flag-staff is \( 75(\sqrt{3} - 1)\text{ m} \) (or approximately \( 54.9\text{ m} \)).
In simple words: The road width of 75 meters and the 45-degree angle mean that the cliff is also 75 meters high. Using the 60-degree angle for the top of the flagstaff, we find that the flagstaff itself is about 54.9 meters tall.

Exam Tip: Be precise about which angle belongs to the bottom and which to the top. The smaller angle (\( 45^\circ \)) always corresponds to the lower point (the top of the cliff), and the larger angle (\( 60^\circ \)) to the higher point (the top of the flag-staff).

Question. The length of the shadow of a tower is √3 times its height. Find the angle of elevation of the Source of light.
Answer: Let \( h \) be the height of the tower. The length of the shadow is \( \sqrt{3} \) times the height, which is \( \sqrt{3}h \). Let \( \theta \) be the angle of elevation of the light source. In the right-angled triangle formed by the tower and its shadow: \( \tan\theta = \frac{\text{Height of the tower}}{\text{Length of the shadow}} \) \( \tan\theta = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}} \) Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( \implies \theta = 30^\circ \) Thus, the angle of elevation of the light source is \( 30^\circ \).
In simple words: Since the shadow is longer than the tower by a factor of \( \sqrt{3} \), the ratio of the height to the shadow length is \( 1/\sqrt{3} \). This represents an elevation angle of exactly 30 degrees.

Exam Tip: For standard questions of this nature, associate the shadow with the adjacent side and the tower height with the opposite side. It is a quick and frequently occurring 1-mark question.

Question. The angle of elevation θ of the top of a light house at a point A on the ground is such that tan θ = 5/12. When the point is moved 240m towards the light house, the angle of elevation becomes φ such that tan φ = 3/4. Find the height of the light house.
Answer: Let \( BC = h \) be the height of the lighthouse. Let \( A \) be the initial point on the ground, and let \( D \) be the new point after moving \( 240\text{ m} \) towards the lighthouse. This gives \( AD = 240\text{ m} \) and \( CD = CA - 240 \). The angles of elevation at \( A \) and \( D \) are \( \theta \) and \( \phi \) respectively. We are given: \( \tan\theta = \frac{5}{12} \) \( \tan\phi = \frac{3}{4} \) In the right-angled triangle \( BCA \): \( \tan\theta = \frac{BC}{CA} \) \( \frac{5}{12} = \frac{h}{CA} \implies CA = \frac{12h}{5} \) - (Equation 1) In the right-angled triangle \( BCD \): \( \tan\phi = \frac{BC}{CD} \) \( \frac{3}{4} = \frac{h}{CD} \implies CD = \frac{4h}{3} \) - (Equation 2) We know that: \( CA - CD = AD \) \( \frac{12h}{5} - \frac{4h}{3} = 240 \) Multiplying the entire equation by 15 (the LCM of 5 and 3) to clear fractions: \( 3(12h) - 5(4h) = 240 \times 15 \) \( 36h - 20h = 3600 \) \( 16h = 3600 \)
\( \implies h = \frac{3600}{16} = 225\text{ m} \) Therefore, the height of the lighthouse is \( 225\text{ m} \).
In simple words: The ratio of height to distance starts at 5/12 and becomes 3/4 after moving 240 meters closer. By setting up the corresponding algebraic equations and solving for height, we find that the lighthouse is 225 meters tall.

Exam Tip: Expressing the horizontal segments in terms of the height \( h \) and working with their difference is the cleanest way to solve this style of problem. Multiplying by the LCM to clear fractions prevents common arithmetic slips.