CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 06

LEVEL-1

Question 1. If the observer is at point B and the object is at point C. What is the angle formed? (Elevation / depression)
Answer: As the observer is at point B and the object is at point C, the observer has to look downwards.
Therefore, the angle formed is the angle of depression.
In simple words: When looking down from a higher point to a lower point, the angle formed with the horizontal is an angle of depression.

Exam Tip: Remember that the angle of elevation and angle of depression are equal in magnitude due to alternate interior angles when horizontal lines are parallel.

Question 2. In the given fig: find BC if AC = 6 cm and \(\angle A = 30^\circ\).
Answer: In right-angled triangle ABC:
\(\angle A = 30^\circ\)
\(\sin A = \frac{BC}{AC}\)
\(\sin 30^\circ = \frac{BC}{6}\)
\(\frac{1}{2} = \frac{BC}{6}\)
\(BC = \frac{6}{2} = 3\) cm
In simple words: Use the sine trigonometric ratio since opposite side and hypotenuse are involved.

Exam Tip: Memorize standard trigonometric table values such as \(\sin 30^\circ = \frac{1}{2}\) to solve quickly.

Question 3. A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be \(60^\circ\). Find the height of the tower.
Answer: Let AB be the tower and C be the point on the ground 20 m away from the foot of the tower.
Angle of elevation \(\angle ACB = 60^\circ\)
Distance \(BC = 20\) m
In right-angled triangle ABC:
\(\frac{AB}{BC} = \tan 60^\circ\)
\(\frac{AB}{20} = \sqrt{3}\)
\(AB = 20\sqrt{3}\) m
Hence, the height of the tower is \(20\sqrt{3}\) m.
In simple words: Use tangent ratio (\(\frac{\text{Perpendicular}}{\text{Base}}\)) when height and adjacent distance are given.

Exam Tip: Always draw a clear rough sketch showing the right-angled triangle with given angles and distances before applying trigonometric ratios.

Question 4. An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from his eye is \(45^\circ\). What is height of the chimney?
Answer: Let the height of the chimney be total height, and let the observer's eye be at point D.
Distance from chimney \(EC = 28.5\) m, so \(DB = 28.5\) m.
Height of observer \(ED = 1.5\) m.
Let height of the chimney above the observer's eye level be \(x\).
In right-angled triangle ADB:
\(\tan 45^\circ = \frac{x}{DB}\)
\(1 = \frac{x}{28.5}\)
\(x = 28.5\) m
Total height of the chimney = \(x + 1.5 = 28.5 + 1.5 = 30\) m.
In simple words: Subtract the observer's height from the total setup, find the upper height using tangent, and then add the observer's height back.

Exam Tip: When observer height is given, never forget to add it to the calculated vertical height to get the total height of the object.

LEVEL - 2

Question 1. From a point P on a level ground, the angle of elevation of the top tower is \(30^\circ\). If the tower is 100 m high, find the distance of point P from the foot of the tower.
Answer: Let AB be the tower such that \(AB = 100\) m.
Let P be the point on the ground, and distance from foot of tower be AP.
\(\angle APB = 30^\circ\)
In right-angled triangle PAB:
\(\frac{AB}{AP} = \tan 30^\circ\)
\(\frac{100}{AP} = \frac{1}{\sqrt{3}}\)
\(AP = 100\sqrt{3}\) m
\(AP = 100 \times 1.73 = 173\) m
In simple words: Divide the height of the tower by \(\tan 30^\circ\) to find the base distance.

Exam Tip: Substitute \(\sqrt{3} = 1.73\) or \(1.732\) only in the final step unless specified otherwise in the question.

Question 2. Find the angle of elevation of the sun, when the length of the shadow of a tree \(\sqrt{3}\) times the height of the tree.
Answer: Let AB be the tree and AC be its shadow.
Let height of the tree \(AB = h\).
Then shadow length \(AC = \sqrt{3}h\).
Let angle of elevation \(\angle ACB = \theta\).
In right-angled triangle ABC:
\(\tan \theta = \frac{AB}{AC}\)
\(\tan \theta = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}}\)
Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), we get \(\theta = 30^\circ\).
In simple words: Ratio of height to shadow gives tangent of the angle, which turns out to be \(30^\circ\).

Exam Tip: In angle problems, equate the ratio and look up standard trigonometric values to find the exact angle.

Question 3. A ladder 10 m long just reaches the top of a wall and makes an angle of \(60^\circ\) with the wall. Find the distance of the foot of the ladder from the wall (\(\sqrt{3} = 1.73\)).
Answer: Let BA be the ladder and AC be the wall.
Length of ladder \(BA = 10\) m.
Angle between ladder and wall \(\angle BAC = 60^\circ\).
Distance of foot of ladder from wall = BC.
In right-angled triangle ABC:
\(\sin 60^\circ = \frac{BC}{BA}\)
\(\frac{\sqrt{3}}{2} = \frac{BC}{10}\)
\(BC = \frac{10\sqrt{3}}{2} = 5\sqrt{3}\)
\(BC = 5 \times 1.73 = 8.65\) m.
In simple words: Use sine function because the angle is given with the vertical wall instead of the ground.

Exam Tip: Be careful with angles given with the vertical (wall) instead of the horizontal (ground); trigonometric ratios swap accordingly (\(\sin\) becomes base/hypotenuse equivalent).

Question 4. The angle of elevation of a ladder leaning against a wall is \(60^\circ\) and the foot of the ladder is 4.6 m away from the wall. Find the length of the ladder.
Answer: Let AB be the wall and BC be the ladder.
Distance from foot of ladder to wall \(AC = 4.6\) m.
Angle of elevation \(\angle ACB = 60^\circ\).
In right-angled triangle ABC:
\(\frac{AC}{BC} = \cos 60^\circ\)
\(\frac{4.6}{BC} = \frac{1}{2}\)
\(BC = 2 \times 4.6 = 9.2\) m.
In simple words: Use cosine ratio since adjacent side (distance from wall) and hypotenuse (ladder length) are involved.

Exam Tip: Cosine relates the adjacent side to the hypotenuse in right triangles.

LEVEL-3

Question 1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships is \(30^\circ\) and \(45^\circ\) respectively. If the lighthouse is 100 m high, find the distance between the two ships.
Answer: Let AB be the lighthouse of height 100 m.
Let C and D be the positions of the two ships on opposite sides.
Given \(AB = 100\) m, \(\angle ACB = 30^\circ\), \(\angle ADB = 45^\circ\).
In right triangle ABC:
\(\frac{AB}{AC} = \tan 30^\circ = \frac{1}{\sqrt{3}}\)
\(AC = AB \times \sqrt{3} = 100\sqrt{3}\) m.
In right triangle ABD:
\(\frac{AB}{AD} = \tan 45^\circ = 1\)
\(AD = AB = 100\) m.
Total distance between the two ships CD = \(AC + AD\)
\(CD = 100\sqrt{3} + 100 = 100(\sqrt{3} + 1)\)
\(CD = 100(1.73 + 1) = 100 \times 2.73 = 273\) m.
In simple words: Calculate distances from both sides separately using right triangles and add them together.

Exam Tip: For objects on opposite sides, add the individual base distances; for objects on the same side, subtract them.

Question 2. On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are \(45^\circ\) and \(60^\circ\). If the height of the tower is 600 m, find the distance between the objects.
Answer: Let DC be the tower of height 600 m.
Let A and B be the two objects on the same side.
Given \(DC = 600\) m, \(\angle DAC = 45^\circ\), \(\angle DBC = 60^\circ\).
In right triangle DBC:
\(\tan 60^\circ = \frac{DC}{BC}\)
\(\sqrt{3} = \frac{600}{BC}\)
\(BC = \frac{600}{\sqrt{3}}\) m.
In right triangle DAC:
\(\tan 45^\circ = \frac{DC}{AC}\)
\(1 = \frac{600}{AC}\)
\(AC = 600\) m.
Distance between the objects AB = \(AC - BC\)
\(AB = 600 - \frac{600}{\sqrt{3}} = 600 - 200\sqrt{3} = 600 - 200(1.732) = 600 - 346.4 = 253.6 \approx 254\) m.
In simple words: Find both base lengths and subtract the smaller base from the larger base to get the distance between the objects.

Exam Tip: Ensure proper rationalization or decimal approximation for radical denominators.

Question 3. The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are \(45^\circ\) and \(60^\circ\). What is the distance between these two points?
Answer: Let BD be the lighthouse of height 60 m.
Let A and C be two points on opposite sides of the lighthouse on the ground.
\(\angle BAD = 45^\circ\), \(\angle BCD = 60^\circ\).
In right triangle ABD:
\(\tan 45^\circ = \frac{BD}{BA}\)
\(1 = \frac{60}{BA}\)
\(BA = 60\) m.
In right triangle CBD:
\(\tan 60^\circ = \frac{BD}{BC}\)
\(\sqrt{3} = \frac{60}{BC}\)
\(BC = \frac{60}{\sqrt{3}} = 20\sqrt{3} = 20 \times 1.73 = 34.6\) m.
Total distance between the two points AC = \(BA + BC\)
\(AC = 60 + 34.6 = 94.6\) m.
In simple words: Compute both bases separately and add them since the points are on opposite sides.

Exam Tip: Double-check arithmetic calculations when substituting decimal values for square roots.

Question 4. A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are \(30^\circ\) and \(60^\circ\) respectively. What is the height of the tower?
Answer: Let DC be the vertical tower of height \(h\), and let AD be the vertical flagpole of height 18 m.
Let B be the point of observation on the ground.
Given \(AD = 18\) m, \(\angle DBC = 30^\circ\), \(\angle ABC = 60^\circ\).
In right triangle DCB:
\(\tan 30^\circ = \frac{DC}{BC}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{BC}\)
\(BC = h\sqrt{3}\) ... (1)
In right triangle ACB:
\(\tan 60^\circ = \frac{AC}{BC}\)
\(\sqrt{3} = \frac{18 + h}{BC}\)
\(BC = \frac{18 + h}{\sqrt{3}}\) ... (2)
Equating (1) and (2):
\(h\sqrt{3} = \frac{18 + h}{\sqrt{3}}\)
\(3h = 18 + h\)
\(2h = 18\)
\(h = 9\) m.
Hence, the height of the tower is 9 m.
In simple words: Equate the common base obtained from both triangles to solve for the height of the tower.

Exam Tip: Equating the shared base expression is a standard technique for multi-part vertical structures.

LEVEL-4

Question 1. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Answer: Let AB be the tower of height \(h\).
Let C and D be two points at distances 4 m and 9 m from the base B respectively, so \(BC = 4\) m and \(BD = 9\) m.
Let \(\angle ACB = \theta\). Then \(\angle ADB = 90^\circ - \theta\) (since they are complementary).
In right triangle ABC:
\(\tan \theta = \frac{AB}{BC} = \frac{h}{4}\) ... (1)
In right triangle ABD:
\(\tan(90^\circ - \theta) = \frac{AB}{BD}\)
\(\cot \theta = \frac{h}{9}\) ... (2)
Multiplying equations (1) and (2):
\(\tan \theta \times \cot \theta = \frac{h}{4} \times \frac{h}{9}\)
\(1 = \frac{h^2}{36}\)
\(h^2 = 36\)
\(h = 6\) m.
Hence, proved that the height of the tower is 6 m.
In simple words: Multiply the tangent and cotangent of complementary angles to eliminate the angle and solve for height.

Exam Tip: Whenever complementary angles are mentioned, use \(\tan \theta\) and \(\cot \theta\) product property equal to 1.

Question 2. The angles of elevation and depression, of an aeroplane and its image in water are \(30^\circ\) and \(45^\circ\), respectively, from the top of a tree of height 5 m. Find the height at which the aeroplane is flying from the ground.
Answer: Let AB be the tree of height 5 m.
Let C be the position of the aeroplane and F be its reflection in water.
Height of aeroplane from ground CE = \(h\).
Distance of reflection from water EF = \(h\).
Let the horizontal distance from the tree to the aeroplane be \(AD = x\).
In right triangle DAF:
\(\tan 45^\circ = \frac{DE + EF}{AD}\)
\(1 = \frac{5 + h}{x}\)
\(x = 5 + h\) ... (1)
In right triangle CAD:
\(\tan 30^\circ = \frac{CD}{AD} = \frac{h - 5}{x}\)
\(\frac{1}{\sqrt{3}} = \frac{h - 5}{x}\)
\(x = \sqrt{3}(h - 5)\) ... (2)
Equating (1) and (2):
\(5 + h = \sqrt{3}(h - 5)\)
\(5 + h = h\sqrt{3} - 5\sqrt{3}\)
\(5 + 5\sqrt{3} = h(\sqrt{3} - 1)\)
\(h = \frac{5(\sqrt{3} + 1)}{\sqrt{3} - 1}\)
Rationalizing the denominator:
\(h = \frac{5(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{5(3 + 1 + 2\sqrt{3})}{3 - 1} = \frac{5(4 + 2\sqrt{3})}{2} = 5(2 + \sqrt{3})\) m.
Hence, the height of the aeroplane from the ground is \(5(2 + \sqrt{3})\) m.
In simple words: Set up equations for horizontal distance using both the aeroplane and its reflection, then equate them and rationalize.

Exam Tip: The distance of an object's reflection below the water surface equals the height of the object above the water surface.

Question 3. 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 \(30^\circ\). The car 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 time taken by the car to reach the foot of the tower from this point.
Answer: Let AB be the tower of height \(h\).
Let C be the initial position of the car and D be the position after 6 seconds.
Let \(CD = x\) and \(DB = y\).
In right triangle ABD:
\(\tan 60^\circ = \frac{AB}{DB}\)
\(\sqrt{3} = \frac{h}{y}\)
\(h = y\sqrt{3}\) ... (1)
In right triangle ABC:
\(\tan 30^\circ = \frac{AB}{CB} = \frac{h}{x + y}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{x + y}\)
\(x + y = h\sqrt{3}\) ... (2)
Substitute \(h = y\sqrt{3}\) into (2):
\(x + y = (y\sqrt{3})\sqrt{3}\)
\(x + y = 3y\)
\(x = 2y\) ... (3)
Since the car covers distance \(x\) in 6 seconds, distance \(2y\) takes 6 seconds.
Therefore, distance \(y\) takes \(\frac{6}{2} = 3\) seconds.
Hence, the car will reach the foot of the tower in 3 seconds.
In simple words: Use the uniform speed proportion to relate distance traveled to the time taken.

Exam Tip: Time taken is directly proportional to distance covered when speed is uniform.

Question 4. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are \(45^\circ\) and \(60^\circ\) respectively. Find height of the tower.
Answer: Let BC be the building of height 20 m, and let AB be the transmission tower of height \(h\) fixed on top of the building.
Let P be a point on the ground such that \(\angle BPC = 45^\circ\) and \(\angle APC = 60^\circ\).
In right triangle PCB:
\(\tan 45^\circ = \frac{BC}{PC}\)
\(1 = \frac{20}{PC}\)
\(PC = 20\) m.
In right triangle PCA:
\(\tan 60^\circ = \frac{AC}{PC} = \frac{h + 20}{20}\)
\(\sqrt{3} = \frac{h + 20}{20}\)
\(h + 20 = 20\sqrt{3}\)
\(h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1)\) m.
Hence, the height of the tower is \(20(\sqrt{3} - 1)\) m.
In simple words: Find the common base distance using the building height, then use it to find the total height and subtract the building height.

Exam Tip: Always find the common base first before evaluating combined heights in multi-tiered structural problems.