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Q1. The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called:

• 1) Vertical angle
• 2) Angle of elevation
• 3) Obtuse angle
• 4) Angle of depression

Solution

The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called angle of elevation.

Q2. Two men on either side of a cliff, 60 m high, observe the angles of elevation of the top of the cliff to be 45^o and 60^o respectively. Find the distance between two men.

Solution

Let A and B represent the position of the two men. In BCD,

Q3. A tree is broken by the wind. The top struck the ground at an angle of 30^o and at a distance of 30 metres from its root. Find the whole height of tree. (Use )

Solution

Let AB be the tree. Suppose the tree is broken at C. OC = CB = y m OA = 30 m (given) In OAC, tan 30^o = or x = 10 m Also, cos 30^o = y = 20 Height of the tree = (x + y) m = 30 m = 51.96 m

Q4. An aricraft is flying at constant height with a speed of 360 km/hours. From a point on the ground, the angle of elevation at an instant was observed to be 45^o. After 20 seconds, the angle of elevation was observed to be 30^o. Determine the height at which the aircraft is flying. (Use = 1.732

Solution

Distance AB = Speed Time = DC = 2 km In right AOD tan 45^o =   In right BOC   Height at which aircraft is flying = km Or 2.732 km

Q5. Two buildings face each other. The height of one building is greater than that of the other. The smaller building is 80 m high. The angles of depression of the top and the bottom of the smaller building from the top of the taller building are 45º and 60º respectively. Find the height of the taller building.

Solution

  In the given figure, PC denotes the taller building and AB denotes the 80 m tall building.     We are interested in determining the height of the taller building, I which is., PC    If we look at the figure carefully and observe that PB is a transversal to the parallel lines PQ and BD.   Therefore, QPB and PBD are alternate angles, and are equal.   So, PBD = 45º. Similarly, PAC = 60º.   In the right angled triangle PBD, we have  

Q6. From an aero- plane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aero-plane are observed to be  and . Show that the height in miles of the aero-plane above the road is given by

Solution

Let AB be the height of the aero-plane = h miles, C and D be the positions of two consecutive mile stones on opposite sides of the aero-plane. Let the distance CB = x, and DB = y From the figure, ACB = , ADB = Adding ( i ) and (ii)      

Q7. A tree casts a shadow 4 m long on the ground, when the angle of elevation of the sun is 45^o. The height of the tree is:

• 1) 3 m
• 2) 5.2 m
• 3) 4 m
• 4) 4.5 m

Solution

Let AB represent the tree and BC be its shadow. Given, BC = 4 m We have:

Q8. A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60^o. Find the distance between the foot of the tower and the ball.

Solution

For correct figure Let C be the point where the ball is. C = 60^o (alternate angles) In ABC, tan 60^o = AB/BC = 20/x x = 20/ = 20 (/3) = 11.53 m

Q9. The fig., shows the observation of point C from point A. The angle of depression from A is:

• 1) 30^o
• 2) 75^o
• 3) 45^o
• 4) 60^o

Solution

Let be the angle of depression. Then, will also be . tan = = tan 30^o Hence, the angle of depression, = 30^o

Q10. If the height and length of the shadow of a man are the same, then the angle of elevation of the sun is

• 1) 45^o
• 2) 15^o
• 3) 60^o
• 4) 30^o

Solution

In the figure, AB is the height of the man and BC is his shadow. Given that, AB = BC In ABC,

Q11. A kite is flying at a height of 90 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60^o. Find the length of the string assuming that there is no slack in the string [Take = 1.732]

Solution

Q12. The angle of depression from the top of a tower 12 m high, at a point on the ground is 30^o. The distance of the point from the top of the tower is:

• 1) 12 m
• 2) 6 m
• 3) 24 m
• 4)

Solution

In the figure, AB denotes the tower of height 12 m. In ABC, we have: Thus, the required distance of the point from the top of the tower is 24 m.

Q13. If sun's elevation is 60^o then a pole of height 6 m will cast a shadow of length

• 1)
• 2)
• 3)
• 4)

Solution

In the figure, AB denotes the pole of height 6 m and OA is its shadow. In OAB, Thus, the length of the shadow of the pole is .

Q14. A tower stands vertically on the ground. From a point on the ground which is 60 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60^o. Find the height of the tower.

Solution

Let h be the height of the tower. In ABC, = tan 60^o h = Height of the tower is m.

Q15. The ratio of the length of rod and its shadow is 1: , then the angle of elevation of the sun is:

• 1) 30^o
• 2) 90^o
• 3) 60^o
• 4) 45^o

Solution

Here AB is the rod whose shadow is BC.

Q16.

Solution

Q17. If the length of the shadow of a pole of height  is 3 m. find the angle of elevation of the sun.

Solution

Let AB be the pole and AC be its shadow. Let the angle of elevation, ACB = θ⁰ Then AB =  and AC = 3 m

Q18. John stands on the bridge and observes that the angles of depression of the banks on opposite sides of river Absyssa are 45º and 60º respectively. The bridge is at a height of 6 m from the banks. Calculate the width of the river Absyssa?

Solution

            In the figure, let A and B represent points on the bank on opposite sides of the river Absyssa, so that AB is its width.   P is a point on the bridge at a height of 6 m, that is, DP = 6 m.   We have to calculate the width of the river, which is the length of the side AB of triangle APB.   Now, AB = AD + DB   In the right angled triangle APD, ÐA = 45º Therefore, we have calculated the width of the river Absyssa to be equal to  m.

Q19. From a point on the ground the angle of elevation of the bottom and the top of a flagstaff situated on the top of a 120 m tall house, was found to be 30^o and 45^o respectively. Find the height of the flagstaff.

Solution

Let AB (= 120 m) be the height of house and MA (= h) be the height of flagstaff. In ABC, tan 30^o = In MCB, Thus, the height of the flagstaff is m.

Q20. From the top of hill the angles of depression of two consecutive kilometer stones due east are found to be 30^o and 60^o. Find the height of the hill.

Solution

For the correct fig. Let PQ be the hill and A and B are the two kilometer stones. PAQ = 60^o, and PBQ = 30^o In PAQ, tan 60^o = PQ/PA = x/y y = x(i) In PBQ tan 30^o = PQ/PB 1/ = x/(y + 1) x = (y + 1)/ (ii) 3y = y + 1 2y = 1 y = ½ km = 500 m x = 500 1.73 = 865 m Height of the hill = 865 m

Q21. From a window, 60 m high above the ground, of a house in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are 60^o and 45^o respectively. Show that the height of the opposite house is 60 (1 +) metres.

Solution

Let E be the window which is 60 m above the ground, AC be the other house. InCDE, y = 60 m InABE, Height of opposite house = (x + 60) m = (60 + 60) m = 60 (1 + ) m

Q22. The angle of elevation of the top of a tower at a point on the ground is 45^o. After going 40 m towards the foot of the tower, the angle of elevation of the top of tower changes to 60^o. Find the height of the tower. (use = 1.73)

Solution

_{Let AB be the tower. Let C and D be the points on the ground. In ABD,} … (1) _{In ABC,} Height of tower = 94.6 m

Q23. From the top of a light house the angle of depression of a ship sailing towards it was found to 30^o. After 10 seconds the angle of depression changes to 60^o. Assuming that the ship is sailing uniform speed, from how much it will take to reach the light house?

Solution

Let AC be the light house of height h. The two positions of the ship are D and B. DB is the distance covered by ship in 10 seconds. In ABC, tan 60^o = … (1) Now in ADC, tan 30^o = DB is twice the distance BC Time taken by ship to cover distance BC will be half as that of DB Time taken by the ship to reach the light house = seconds = 5 seconds

Q24. An electric pole is 10 m high. If its shadow is m in length. Find the angle of elevation of the sun at that time.

Solution

In the figure, AB is the tower and BC denotes its shadow. Let be angle of elevation of the sun. tan = Thus, the angle of elevation of the sun is 30^o.

Q25. Find the height of a mountain if the elevation of its top at an unknown distance from the base is 60^o and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 30^o.

Solution

Let h be the height of the mountain. In CAB, Or h = x ...(i) In DAB, Or [Using (i)] 3x = 10 + x Or x = 5 h = = 5 Thus, the height of the mountain is 5 _km.

Q26. A man on the top of a vertical tower observes a car moving towards the tower. If it takes 12 minutes for the angle of depression to change from 30^o to 45^o, how soon after this car will reach the tower? (see figure)

Solution

Let the speed of the car be x units/min. Then, DC = 12x Let the height of the tower be h units. We have: ACB = 45^o and ADB = 30^o In right triangle ABC, tan 45^o = h = BC In right triangle ABD, tan 30^o = Required time taken by the car to reach the tower = = = = 6(1.732 + 1) = 16.392 minutes

Q27. A circus artist is climbing a rope 12 m long which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground is 30^o.

Solution

In the figure, AB represents the pole and C is the point on the ground. Let the height of the pole be h. In sin 30^o = Thus, the height of the pole is 6 m.

Q28. An aeroplane at an altitude of 200 m observes the angles of depression of two opposite points on two banks of the river to be 45^o and 60^o. Find in metres, the width of the river. (use = 1.732)

Solution

In the figure, C denotes the position of the aeroplane. Points A and B denotes the position of the two points on two banks of the river. We have to find AB, i.e., a + b In ACD, In BCD, Width of the river = 200 + = = 315.5 m

Q29. A boy 2 m tall is standing at some distance from a 30 m tall building. The angle of elevation from his eyes of the top of the building increases form 30^o to 60^o as he walks towards the building. Find the distance he walked towards the building.

Solution

In AGF, tan 60^o= In AGE, tan 30^o = Distance walked = 18.67m

Q30.

Solution

Q31. Distance between two buildings is 130 m. The angle of elevation of the top of the first building when seen from the second building is 45^o. If the height of the second building is 70 m, find the height of the first building.

Solution

Let AB and CD be the two buildings of height h m and 60 m respectively. AC = 130 m. The angle of elevation of top of AB as seen from top of CD is 45^o In triangle DEB,  BE = 130 m AB = AE + BE = CD + BE = 70 + 130 = 200 m. Hence the height of the second building = 200 m.

Q32. A ladder is leaning against a wall making an angle of 60⁰ with the ground. Find the distance of the foot of the ladder from the wall if the other end is 14 m above the ground?

Solution

Let AB be the wall and BC be the ladder. angle of elevation, ∠ACB = 60⁰ Then AB = 14 m and let AC = x m   Distance of the foot of the ladder from the wall is 7 m.

Q33.

Solution

Q34. A tree is broken by the wind. The top struck the ground at an angle of 30^o and at a distance of 30 metres from the root. Find the whole height of the tree.

Solution

Let BD be the whole height of the tree The tree breaks at point C and makes  as given into the figure such that AC = CD     In rt.          Also  Total height of tree = BC + AC