CBSE Class 10 Mathematics Trigonometry Assignment Set 20

Question. The length of a string between a kite and a point on the ground is 150 m. If the string makes an angle \(\theta\) with the ground level such that \(\tan\theta = \frac{8}{15}\), find the height of the kite from the ground.
Answer:
Let \(AB\) be the vertical height of the kite from the ground and \(AC = 150\text{ m}\) be the length of the string. The angle of elevation is \(\theta\). In the right-angled triangle \(ABC\): \[ \frac{AB}{AC} = \sin\theta \implies \frac{AB}{150} = \sin\theta \] We are given: \[ \tan\theta = \frac{8}{15} \] Since \(\tan\theta = \frac{\text{Perpendicular}}{\text{Base}}\), we can find the hypotenuse using the Pythagorean theorem: \[ \text{Hypotenuse} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \] Using this hypotenuse, we calculate \(\sin\theta\):
\( \implies \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{8}{17} \) Now, equating both expressions for \(\sin\theta\): \[ \frac{8}{17} = \frac{AB}{150} \]
\( \implies AB = \frac{8 \times 150}{17} \]
\( \implies AB = \frac{1200}{17} \approx 70.6\text{ m} \) Hence, the height of the kite is approximately 70.6 m.
In simple words: We first find the sine of the angle by calculating the hypotenuse of the triangle from the given tangent. Then, we use this sine ratio to solve for the vertical height of the kite.
Exam Tip: Identifying standard Pythagorean triples like (8, 15, 17) allows you to transition between trigonometric ratios quickly and accurately without long calculations.

Question. A flagstaff stands on the top of a cliff. From a point on the ground 75 m away from the base of the cliff, the angles of elevation of the top and bottom of the flagstaff are \(60^\circ\) and \(45^\circ\) respectively. Find the height of the flagstaff.
Answer:
Let \(AB\) represent the height of the cliff and \(AC\) be the height of the flagstaff. Let \(D\) be the point on the ground at a distance of \(BD = 75\text{ m}\) from the base. In the right-angled triangle \(ABD\): \[ \frac{AB}{BD} = \tan 45^\circ \implies \frac{AB}{75} = 1 \]
\( \implies AB = 75\text{ m} \) -- (1) In the right-angled triangle \(CBD\): \[ \frac{BC}{BD} = \tan 60^\circ \implies \frac{BC}{75} = \sqrt{3} \]
\( \implies BC = 75\sqrt{3}\text{ m} \) -- (2) The height of the flagstaff is \(AC = BC - AB\): \[ AC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)\text{ m} \] Using \(\sqrt{3} \approx 1.732\):
\( \implies AC \approx 75(1.732 - 1) = 75(0.732) = 54.9\text{ m} \) Therefore, the height of the flagstaff is 54.9 m.
In simple words: We find the heights of both the top of the flagpole and the top of the cliff from the ground. Subtracting the cliff's height from the overall height gives the height of the flagpole.
Exam Tip: Clearly write down intermediate equations and state the values used for trigonometric functions (such as \(\tan 60^\circ = \sqrt{3}\)) to receive step-wise marks.

Question. Find the angle of elevation of the sun when the length of the shadow of a vertical pole is \(\sqrt{3}\) times the height of the pole.
Answer:
Let \(BC = h\) be the height of the vertical pole and \(AB = \sqrt{3}h\) be the length of its shadow. Let \(\theta\) be the angle of elevation of the sun. In the right-angled triangle \(ABC\): \[ \tan\theta = \frac{BC}{AB} \]
\( \implies \tan\theta = \frac{h}{\sqrt{3}h} \]
\( \implies \tan\theta = \frac{1}{\sqrt{3}} \] We know that \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\( \implies \tan\theta = \tan 30^\circ \)
\( \implies \theta = 30^\circ \) Hence, the angle of elevation of the sun is \(30^\circ\).
In simple words: The ratio of the pole's height to its shadow's length gives the tangent of the angle. Since this ratio equals one over the square root of three, the angle is thirty degrees.
Exam Tip: Be sure to write the formula \(\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}\) explicitly before substituting values to make your working clear and logical.

Question. From two points on the ground on the same side of a vertical tower, the angles of elevation of the top of the tower are \(\phi\) and \(\theta\) respectively. The distance between the two points is 240 m. If \(\tan\phi = \frac{3}{4}\) and \(\tan\theta = \frac{5}{12}\), calculate the height of the tower.
Answer:
Let \(AB = h\) represent the height of the tower. Let \(C\) and \(D\) be the two points on the ground such that \(CD = 240\text{ m}\). In the right-angled triangle \(ABC\): \[ \frac{h}{BC} = \tan\phi = \frac{3}{4} \implies BC = \frac{4h}{3} \] In the right-angled triangle \(ABD\): \[ \frac{h}{BD} = \tan\theta = \frac{5}{12} \implies BD = \frac{12h}{5} \] Since \(BD - BC = DC = 240\text{ m}\):
\( \implies \frac{12h}{5} - \frac{4h}{3} = 240 \) Finding a common denominator of 15:
\( \implies \frac{36h - 20h}{15} = 240 \)
\( \implies \frac{16h}{15} = 240 \)
\( \implies 16h = 240 \times 15 \)
\( \implies h = \frac{3600}{16} = 225\text{ m} \) Hence, the height of the tower is 225 m.
In simple words: We write the horizontal distances from both points to the tower in terms of its height. Subtracting these distances gives us an equation that helps us solve for the tower's height.
Exam Tip: Be meticulous with algebraic equations involving fractions. Finding the correct lowest common multiple (LCM) for the denominators prevents calculation errors.

Question. From a point on the ground, the angles of elevation of the top of a building and a helicopter hovering at a height of 10 m directly above the top of the building are \(45^\circ\) and \(60^\circ\) respectively. Find the height of the building.
Answer:
Let \(AB = h\) be the height of the building and \(C\) be the position of the helicopter, so \(AC = 10\text{ m}\) and the overall height \(BC = h + 10\text{ m}\). Let \(D\) be the point of observation. In the right-angled triangle \(ABD\): \[ \frac{h}{BD} = \tan 45^\circ = 1 \implies BD = h \] -- (1) In the right-angled triangle \(CBD\): \[ \frac{h+10}{BD} = \tan 60^\circ = \sqrt{3} \] -- (2) Substituting equation (1) into equation (2):
\( \implies \frac{h+10}{h} = \sqrt{3} \)
\( \implies h + 10 = \sqrt{3}h \)
\( \implies \sqrt{3}h - h = 10 \)
\( \implies h(\sqrt{3} - 1) = 10 \)
\( \implies h = \frac{10}{\sqrt{3} - 1} \) Rationalizing the denominator:
\( \implies h = \frac{10(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10(\sqrt{3} + 1)}{3 - 1} = 5(\sqrt{3} + 1)\text{ m} \) Using \(\sqrt{3} \approx 1.732\):
\( \implies h \approx 5(1.732 + 1) = 5(2.732) = 13.66\text{ m} \) Therefore, the height of the building is approximately 13.66 m.
In simple words: We relate the building and helicopter heights to the horizontal ground distance. Combining these relations gives a linear equation that lets us calculate the building's height.
Exam Tip: Rationalization of the denominator is highly recommended in such answers. Keeping your steps complete up to two decimal places ensures maximum marks.

Question. Two vertical poles of equal height 50 m stand on a horizontal plane. From a point on the ground, on the line joining their feet and outside the segment joining them, the angles of elevation of the tops of the poles are \(30^\circ\) and \(60^\circ\). Find the distance between the poles.
Answer:
Let \(AD\) and \(BC\) be the two vertical poles of equal height, so \(AD = BC = 50\text{ m}\). Let \(E\) be the point of observation on the horizontal ground. The distance between the poles is \(DC = EC - ED\) -- (1) In the right-angled triangle \(BCE\): \[ \frac{BC}{EC} = \tan 30^\circ \implies \frac{50}{EC} = \frac{1}{\sqrt{3}} \]
\( \implies EC = 50\sqrt{3}\text{ m} \) -- (2) In the right-angled triangle \(ADE\): \[ \frac{AD}{ED} = \tan 60^\circ \implies \frac{50}{ED} = \sqrt{3} \]
\( \implies ED = \frac{50}{\sqrt{3}}\text{ m} \) -- (3) Substituting equations (2) and (3) into equation (1):
\( \implies DC = 50\sqrt{3} - \frac{50}{\sqrt{3}} \)
\( \implies DC = \frac{150 - 50}{\sqrt{3}} = \frac{100}{\sqrt{3}}\text{ m} \) Rationalizing the denominator:
\( \implies DC = \frac{100\sqrt{3}}{3}\text{ m} \approx \frac{100 \times 1.732}{3} \approx 57.7\text{ m} \) Hence, the distance between the two poles is approximately 57.7 m.
In simple words: We calculate the horizontal distances from the observer to each pole using the given angles. Subtracting the smaller distance from the larger distance gives the space between the poles.
Exam Tip: Retain radical forms like \(\sqrt{3}\) through your calculations and perform the final decimal rounding at the very end to avoid approximation slip-ups.

Question. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of \(30^\circ\), 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 \(60^\circ\). Find the total time taken by the car to reach the foot of the tower from the starting point.
Answer:
Let \(AB = h\) represent the height of the tower. Let \(D\) be the initial position of the car and \(C\) be its position after 6 seconds. First, we find the horizontal distances in terms of the height \(h\). In the right-angled triangle \(ABC\): \[ \frac{h}{BC} = \tan 60^\circ = \sqrt{3} \implies BC = \frac{h}{\sqrt{3}} \] In the right-angled triangle \(ABD\): \[ \frac{h}{BD} = \tan 30^\circ = \frac{1}{\sqrt{3}} \implies BD = h\sqrt{3} \] The distance \(DC\) covered in 6 seconds is: \[ DC = BD - BC = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \] Since the car travels at a uniform speed, the time taken is directly proportional to the distance: - To cover a distance of \(\frac{2h}{\sqrt{3}}\), the time taken is 6 seconds. - Therefore, to cover the entire distance \(BD = h\sqrt{3}\), the total time taken is:
\( \implies \text{Total Time} = \frac{6}{\left(\frac{2h}{\sqrt{3}}\right)} \times h\sqrt{3} \)
\( \implies \text{Total Time} = \frac{6\sqrt{3}}{2h} \times h\sqrt{3} = 3 \times 3 = 9\text{ seconds} \) Thus, the total time taken by the car to reach the foot of the tower is 9 seconds.
In simple words: The car travels two-thirds of the total distance in six seconds. Moving at a steady speed, it will take nine seconds in total to reach the tower's base.
Exam Tip: Be careful to read what the question asks for — whether it requires the "remaining time" (which is 3 seconds) or the "total time" (which is 9 seconds).

Question. From the top of a 10 m high building, the angle of elevation of the top of a cable tower is \(60^\circ\) and the angle of depression of its foot is \(45^\circ\). Determine the height of the cable tower.
Answer:
Let \(AB = 10\text{ m}\) be the building and \(CD = h\) be the height of the cable tower. Let \(AE\) be the horizontal line from \(A\) to \(CD\). In the right-angled triangle \(ABD\): \[ \frac{AB}{BD} = \tan 45^\circ \implies \frac{10}{BD} = 1 \]
\( \implies BD = 10\text{ m} = AE \) In the right-angled triangle \(CAE\), the vertical height is \(CE = h - 10\): \[ \frac{CE}{AE} = \tan 60^\circ \implies \frac{h - 10}{10} = \sqrt{3} \]
\( \implies h - 10 = 10\sqrt{3} \]
\( \implies h = 10\sqrt{3} + 10 = 10(\sqrt{3} + 1)\text{ m} \) Using \(\sqrt{3} \approx 1.732\):
\( \implies h \approx 10(1.732 + 1) = 10(2.732) = 27.32\text{ m} \) Hence, the height of the cable tower is 27.32 m.
In simple words: The 45-degree angle of depression means the distance between the structures matches the building's height of 10 meters. Using this width with the 60-degree angle of elevation gives the remaining height of the tower.
Exam Tip: Clearly write down that the alternate interior angles are equal (such as \(\angle ADB = \angle EAD = 45^\circ\)) to justify your steps in the bottom triangle.

Question. A flagpole of height \(h\) stands on top of a tower. From a point on the ground, the angles of elevation of the bottom and top of the flagpole are \(\alpha\) and \(\beta\) respectively. Prove that the height of the tower is \(\frac{h\tan\alpha}{\tan\beta - \tan\alpha}\).
Answer:
Let \(BC = x\) be the height of the tower and \(AB = h\) be the height of the flagpole. Let \(D\) be the point of observation on the ground. In the right-angled triangle \(BCD\): \[ \frac{BC}{DC} = \tan\alpha \implies \frac{x}{DC} = \tan\alpha \] -- (1) In the right-angled triangle \(ACD\): \[ \frac{AC}{DC} = \tan\beta \implies \frac{x + h}{DC} = \tan\beta \] -- (2) Dividing equation (2) by equation (1):
\( \implies \frac{x+h}{x} = \frac{\tan\beta}{\tan\alpha} \)
\( \implies 1 + \frac{h}{x} = \frac{\tan\beta}{\tan\alpha} \)
\( \implies \frac{h}{x} = \frac{\tan\beta}{\tan\alpha} - 1 \)
\( \implies \frac{h}{x} = \frac{\tan\beta - \tan\alpha}{\tan\alpha} \) Taking the reciprocal:
\( \implies \frac{x}{h} = \frac{\tan\alpha}{\tan\beta - \tan\alpha} \)
\( \implies x = \frac{h\tan\alpha}{\tan\beta - \tan\alpha} \) Hence proven.
In simple words: Expressing the horizontal ground distance using both angles of elevation creates an algebraic ratio. Solving this ratio for the tower's height yields the required formula.
Exam Tip: For proof-based questions, write each step clearly and keep the final expression boxed to show completion.

Question. From the top of a building of height \(h\), the angle of elevation of the top of a tower is \(\theta\), and from the foot of the building, the angle of elevation of the top of the tower is \(\phi\). Show that the height of the tower is \(\frac{h\tan\phi}{\tan\phi - \tan\theta}\).
Answer:
Let \(AB = x\) represent the height of the tower and \(CD = h\) be the height of the building. Draw a horizontal line \(DE\) from \(D\) to \(AB\) meeting it at \(E\), so \(EB = CD = h\) and \(AE = x - h\). In the right-angled triangle \(ABC\): \[ \frac{AB}{CB} = \tan\phi \implies \frac{x}{CB} = \tan\phi \]
\( \implies CB = \frac{x}{\tan\phi} \) -- (1) Since \(DE = CB\), we have \(DE = \frac{x}{\tan\phi}\). In the right-angled triangle \(ADE\): \[ \frac{AE}{DE} = \tan\theta \implies \frac{x - h}{\left(\frac{x}{\tan\phi}\right)} = \tan\theta \]
\( \implies (x - h)\tan\phi = x\tan\theta \)
\( \implies x\tan\phi - h\tan\phi = x\tan\theta \) Rearranging to isolate \(x\):
\( \implies x\tan\phi - x\tan\theta = h\tan\phi \)
\( \implies x(\tan\phi - \tan\theta) = h\tan\phi \)
\( \implies x = \frac{h\tan\phi}{\tan\phi - \tan\theta} \) Hence proven.
In simple words: We find the ground width using the angle from the bottom of the building. Applying this same width to the upper triangle sets up a relation that determines the total tower height.
Exam Tip: Be mindful of assigning angles correctly. The larger angle \(\phi\) corresponds to the bottom of the building, while the smaller angle \(\theta\) corresponds to the top.