Samacheer Kalvi Class 10 Maths Solutions Chapter 6 Trigonometry Exercise 6.2

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1. Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height \( 10 \sqrt{3} \) m.
Answer: Let AC be the height of the tower and AB be the distance from the foot of the tower to the observation point on the ground.Height of the tower \( AC = 10 \sqrt{3} \) m.
Distance from the base of the tower to the observation point \( AB = 30 \) m.
Let the angle of elevation \( \angle ABC \) be \( \theta \).
In the right-angled triangle ABC:
\( \tan \theta = \frac{AC}{AB} \)
\( \tan \theta = \frac{10 \sqrt{3}}{30} \)
\( \tan \theta = \frac{\sqrt{3}}{3} \)
\( \tan \theta = \frac{1}{\sqrt{3}} \)
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we have \( \theta = 30^\circ \). The angle of elevation is the angle between the horizontal ground and the line of sight to the top of the tower.
In simple words: The angle from the ground up to the top of the tower is 30 degrees. We used the tangent function, which relates the height of the tower to how far away you are standing.

🎯 Exam Tip: Always draw a clear diagram for trigonometry problems to correctly identify the sides and angles, and remember common tangent values like \( \tan 30^\circ = 1/\sqrt{3} \).

Question 2. A road is flanked on either side by continuous rows of houses of height \( 4\sqrt{3} \) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer: Let AD be the height of the house, and P be the position of the pedestrian on the median of the road. The median divides the road into two equal halves.
Let the midpoint of the road AB be "P", such that \( PA = PB \).
Height of the house \( AD = 4\sqrt{3} \) m.
Let the distance between the pedestrian and the house (AP) be "x".
In the right-angled triangle APD:
\( \tan 30^\circ = \frac{AD}{AP} \)
\( \frac{1}{\sqrt{3}} = \frac{4\sqrt{3}}{x} \)
Now, we solve for x:
\( x = 4\sqrt{3} \times \sqrt{3} \)
\( x = 4 \times 3 \)
\( x = 12 \) m
The width of the road is the total distance from one house to the other, which is \( PA + PB \). Since P is the median, \( PA = PB = x \).
So, the width of the road \( = x + x = 12 + 12 = 24 \) m. The distance from the pedestrian to the house is half the total road width.
In simple words: The distance from the person to one house is 12 meters. Since the person is in the middle of the road, the road is twice that length. So, the total width of the road is 24 meters.

🎯 Exam Tip: Remember that "median of the road" implies the pedestrian is exactly in the middle, making the distance to each side equal.

Question 3. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\( \sqrt{3} = 1.732 \))
Answer: Let D be the position of the man (observer). C is the base of the wall. F is the bottom of the window and E is the top of the window.
Let the height of the window FE be "h" m.
Let the height from the ground to the bottom of the window (FC) be "x" m.
So, the total height to the top of the window (EC) is \( (h+x) \) m.
The man (observer) is 5 m away from the wall (CD = 5 m).

In the right-angled triangle CDF:
\( \tan 45^\circ = \frac{CF}{CD} \)
\( 1 = \frac{x}{5} \)
\( x = 5 \) m

In the right-angled triangle CDE:
\( \tan 60^\circ = \frac{CE}{CD} \)
\( \sqrt{3} = \frac{x+h}{5} \)
\( x+h = 5\sqrt{3} \)
Now, substitute the value of \( x = 5 \) into the equation:
\( 5+h = 5\sqrt{3} \)
\( h = 5\sqrt{3} - 5 \)
\( h = 5 \times 1.732 - 5 \)
\( h = 8.66 - 5 \)
\( h = 3.66 \) m
The height of the window is 3.66 m. Understanding how angles of elevation change with distance helps in many real-world measurements.
In simple words: The window is 3.66 meters tall. We used trigonometry to find its height, first finding the height to the bottom of the window, then to the top, and subtracting them.

🎯 Exam Tip: When dealing with two angles of elevation to different points on the same vertical object, use two separate right-angled triangles and a common horizontal distance for calculations.

Question 4. stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal. (\( \tan 40^\circ = 0.8391 \), \( \sqrt{3} = 1.732 \))
Answer: Let AC be the height of the pedestal and CD be the height of the statue. Let AB be the horizontal distance from the observation point to the base of the pedestal.
Height of the statue \( CD = 1.6 \) m.
Let the height of the pedestal \( AC = h \) m.
The total height from the ground to the top of the statue \( AD = h + 1.6 \) m.
Let the horizontal distance from the observation point to the pedestal \( AB = x \) m.

In the right-angled triangle ABD (using the top of the statue):
\( \tan 60^\circ = \frac{AD}{AB} \)
\( \sqrt{3} = \frac{h+1.6}{x} \)
From this, we get \( x = \frac{h+1.6}{\sqrt{3}} \) .......(1)

In the right-angled triangle ABC (using the top of the pedestal):
\( \tan 40^\circ = \frac{AC}{AB} \)
\( 0.8391 = \frac{h}{x} \)
From this, we get \( x = \frac{h}{0.8391} \) .......(2)

Now, substitute the value of x from equation (2) into equation (1):
\( \frac{h}{0.8391} = \frac{h+1.6}{\sqrt{3}} \)
\( h \times \sqrt{3} = (h+1.6) \times 0.8391 \)
\( 1.732h = 0.8391h + 1.6 \times 0.8391 \)
\( 1.732h = 0.8391h + 1.34256 \)
\( 1.732h - 0.8391h = 1.34256 \)
\( 0.8929h = 1.34256 \)
\( h = \frac{1.34256}{0.8929} \)
\( h \approx 1.5035 \) m
Rounding to one decimal place, the height of the pedestal is approximately 1.5 m. This problem shows how comparing angles of elevation helps find unknown heights.
In simple words: The pedestal is about 1.5 meters tall. We used the different angles a person saw to the top of the pedestal and the top of the statue to figure out the pedestal's height.

🎯 Exam Tip: When dealing with two angles from the same observation point, set up two tangent equations and then solve them simultaneously, often by equating the common unknown (like 'x' here).

Question 5. A Flag pole 'h' metres is on the top of the hemispherical dome of radius 'r' metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\( \sqrt{3} = 1.732 \))
Answer: Let A be the center of the base of the hemispherical dome. Let AD be the radius of the dome (r) and D be the top of the dome. Let DE be the height of the flag pole (h). So AE = r+h.
The man is standing 7 m away from the dome. His distance from the center A is \( AF = r+7 \) m. From this point, the angle of elevation to the top of the pole (E) is \( 45^\circ \).
Then he moves 5 m further away. His new distance from the center A is \( AB = r+7+5 = r+12 \) m. From this point, the angle of elevation to the bottom of the pole (D) is \( 30^\circ \).

(ii) First, let's find the radius of the dome (r).
In the right-angled triangle ABD (using the bottom of the pole and the further observation point):
\( \tan 30^\circ = \frac{AD}{AB} \)
\( \frac{1}{\sqrt{3}} = \frac{r}{r+12} \)
\( r+12 = \sqrt{3}r \)
\( 12 = \sqrt{3}r - r \)
\( 12 = r(\sqrt{3} - 1) \)
\( r = \frac{12}{\sqrt{3} - 1} \)
\( r = \frac{12}{1.732 - 1} \)
\( r = \frac{12}{0.732} \)
\( r \approx 16.39 \) m
The radius of the dome is approximately 16.39 m.

(i) Now, let's find the height of the pole (h).
In the right-angled triangle AFE (using the top of the pole and the closer observation point):
\( \tan 45^\circ = \frac{AE}{AF} \)
\( 1 = \frac{r+h}{r+7} \)
\( r+7 = r+h \)
\( h = 7 \) m
The height of the pole is 7 m. These calculations demonstrate how two observations from different points can determine multiple unknown dimensions.
In simple words: The flag pole is 7 meters high, and the round dome it sits on has a radius of about 16.39 meters. We solved this by using the angles from two different viewing spots.

🎯 Exam Tip: For problems involving movement of the observer, clearly label the distances from a fixed point and set up separate trigonometric equations for each observation point, then solve them simultaneously.

Question 6. The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer: Let A be the point of observation on the ground. Let AB be the horizontal distance to the base of the tower and pole. Let BC be the height of the tower (15 m). Let AD be the height of the electric pole (h m). E is the point on the tower at the same height as the top of the pole D.
Let the height of the electric pole AD be "h" m (meaning the pole is of height h and its base is on the ground at point B).
So, from the ground, the height of the pole's top is BE = h.
The top of the tower is C, with height BC = 15 m.
Thus, the part of the tower above the pole's height is \( EC = BC - BE = 15 - h \) m.
Let AB be the horizontal distance between the observer and the tower/pole, which is "x". Also, \( DE = x \).

In the right-angled triangle ABC (from observer A to top of tower C):
\( \tan 60^\circ = \frac{BC}{AB} \)
\( \sqrt{3} = \frac{15}{x} \)
\( x = \frac{15}{\sqrt{3}} \)
\( x = \frac{15\sqrt{3}}{3} \)
\( x = 5\sqrt{3} \) m

In the right-angled triangle CDE (from top of tower C to top of pole D):
\( \tan 30^\circ = \frac{EC}{DE} \)
\( \frac{1}{\sqrt{3}} = \frac{15-h}{x} \) .......(1)

Substitute the value of \( x = 5\sqrt{3} \) into equation (1):
\( \frac{1}{\sqrt{3}} = \frac{15-h}{5\sqrt{3}} \)
Multiply both sides by \( 5\sqrt{3} \):
\( 5 = 15 - h \)
\( h = 15 - 5 \)
\( h = 10 \) m
The height of the electric pole is 10 m. This calculation shows how angles of elevation and depression can be combined to find unknown heights.
In simple words: The electric pole is 10 meters tall. We used the angle to the top of the tower to find the distance, and then used that distance with the angle from the top of the tower to the pole's top to find the pole's height.

🎯 Exam Tip: Carefully identify the observer's position and the reference point for each angle (e.g., from ground to top of tower, from top of tower to top of pole). This prevents mixing up heights and distances.

Question 7. A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer: Let AC be the vertical pole, with C as its base on the ground. Let B be the mark on the pole that divides it into a 1:9 ratio. Since the lower part is shorter, \( BC:AB = 1:9 \).
Let \( BC = x \) and \( AB = 9x \). So the total height of the pole \( AC = x + 9x = 10x \).
D is a point on the ground 25 m away from the base of the pole, so \( CD = 25 \) m.
The problem states that the two parts subtend equal angles at D, meaning \( \angle CDB = \angle BDA \). This indicates that BD is the angle bisector of \( \angle ADC \).

By the Angle Bisector Theorem in \( \triangle ADC \):
\( \frac{AB}{BC} = \frac{AD}{DC} \)
\( \frac{9x}{x} = \frac{AD}{25} \)
\( 9 = \frac{AD}{25} \)
\( AD = 9 \times 25 \)
\( AD = 225 \) m

Now, in the right-angled triangle ACD:
By Pythagoras theorem, \( AD^2 = AC^2 + CD^2 \)
\( (225)^2 = (10x)^2 + (25)^2 \)
\( 50625 = 100x^2 + 625 \)
\( 100x^2 = 50625 - 625 \)
\( 100x^2 = 50000 \)
\( x^2 = \frac{50000}{100} \)
\( x^2 = 500 \)
\( x = \sqrt{500} \)
\( x = \sqrt{100 \times 5} \)
\( x = 10\sqrt{5} \) m

The total height of the pole \( AC = 10x \).
\( AC = 10 \times (10\sqrt{5}) \)
\( AC = 100\sqrt{5} \) m.
The angle bisector theorem is a powerful tool in geometry, connecting side lengths to angle divisions.
In simple words: The pole is \( 100\sqrt{5} \) meters tall. We used a special rule for angles to find the length of one side, then used the Pythagorean theorem to find 'x', and finally the total height.

🎯 Exam Tip: Recognize when the Angle Bisector Theorem can be applied. It's useful when a line segment divides an angle in a triangle, relating the ratios of the opposite sides.

Question 8. A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (\( \tan 4^\circ = 0.0699 \), \( \tan 8^\circ = 0.1405 \))
Answer: Let AC be the height of the mountain peak (h miles). Let A be the base of the mountain.
Let the height of the peak \( AC = h \) mile.
Let the distance from the first milestone (D) to the base of the peak (A) be \( AD = x \) mile.
The second milestone (B) is 1 mile away from the first. So the distance from the second milestone to the base of the peak is \( AB = x + 1 \) mile.

In the right-angled triangle ADC (from milestone D to peak C):
\( \tan 8^\circ = \frac{AC}{AD} \)
\( 0.1405 = \frac{h}{x} \)
So, \( x = \frac{h}{0.1405} \) .......(1)

In the right-angled triangle ABC (from milestone B to peak C):
\( \tan 4^\circ = \frac{AC}{AB} \)
\( 0.0699 = \frac{h}{x+1} \)
So, \( h = 0.0699(x+1) \)
\( h = 0.0699x + 0.0699 \)
\( h - 0.0699 = 0.0699x \)
\( x = \frac{h - 0.0699}{0.0699} \) .......(2)

Now, equate the expressions for x from equations (1) and (2):
\( \frac{h}{0.1405} = \frac{h - 0.0699}{0.0699} \)
\( 0.0699h = 0.1405(h - 0.0699) \)
\( 0.0699h = 0.1405h - 0.1405 \times 0.0699 \)
\( 0.0699h = 0.1405h - 0.00982195 \)
\( 0.00982195 = 0.1405h - 0.0699h \)
\( 0.00982195 = 0.0706h \)
\( h = \frac{0.00982195}{0.0706} \)
\( h \approx 0.1391 \) mile
The height of the peak is approximately 0.14 mile. This method, using two observation points, is common for measuring inaccessible heights.
In simple words: The mountain peak is about 0.14 miles high. We found this by taking two measurements from different places on the road and using the angles to calculate the height.

🎯 Exam Tip: Problems involving two observation points and an unknown height often require setting up two equations and solving them simultaneously to find both the horizontal distance and the vertical height.