Students and Teacher's Forum

Students and Teachers Forum

1. In ∆ABC, if b = 10 cm, c = 12 cm and ∠BAC = 60^o, find the area of ∆ABC.

Given, b = 10 cm c = 12 cm and ∠BAC = 60o We know, Area of ∆ABC = ½ bc SinA .....

2. From the roof of a house, an observer observed the top of a temple. The angle of elevation of the top of the temple as seen from the roof of the house is 45^o. If the height of the house and the temple are 10.5 m and 26.75 m respectively, find the distance between the house and the temple.

Let, AD = height of the house = 10.5 m CE = height of the temple = 26.75 m AB = DE = distance between the house and the temple ∠BAC = 45o In right .....

3. Krishna, a young boy of 1.72 m, tall observes the top of a column and finds an angle of elevation to be 60^o. If the distance between the man and the column is 250 m, find the height of the column.

Let, AD = height of Krishna = 1.72 m AB = DE = distance between the boy and the column = 250 m ∠BAC = 60o In right angled triangle ABC, Tan 60o = .....

4. A circular pond has a pole erected vertically at its centre. The top of the pole is 16√3 m above the surface and the angle of elevation at a point on the circumference is 30^o. Find the radius of the pond.

Let, AB = radius of the pond AC = height of the pole above the surface = 16√3 m ∠ABC = 30o In right angled triangle ABC, Tan 30o = ACAB Or, .....

5. A kite is flying at a height of 75 m from the ground. If the length of the chord paid out at the time of flying the kite is 150 m, find the angle of elevation of the kite.

Let, AC = length of the chord = 150 m BC = distance of the kite from the ground = 75 m ∠CAB = angle of elevation = θ In right angled triangle ABC, Sin .....

6. An eagle flying in the sky 180 m, above the ground notices a fish in a pond at an angle of depression of 60^o. Find the distance of the position of the fish from the vertical line through the position of the eagle.

Let, AB = height of the eagle from the ground = 180 m C is the position of the fish BC = distance of the fish through the position of the eagle. .....

7. In ∆ABC, if AB = 26 cm, BC = 15 cm and AC = 13 cm, find the area of ∆ABC.

Given, AB (c) = 26 cm BC (a) = 15 cm and AC (b) = 13 cm. Now, Semi-perimeter of ∆ABC (s) = 12(a + b + c) .....

8. In ∆ABC, if AB = 12 cm, BC = 10 cm and AC = 8 cm, find the area of ∆ABC.

Solution: Given, AB (c) = 12 cm BC (a) = 10 cm and AC (b) = 8 cm. Now, Semi-perimeter of ∆ABC (s) = 12(a+b+c) .....

9. In ∆ABC, if a = 5 cm, b = 7 cm and c = 8 cm, find the area of ∆ABC.

Given, a = 5 cm b = 7 cm and c = 8 cm Now, Semi-perimeter of ∆ABC (s) = 12 (a+b+c) .....

10. The angle of elevation of the top of a tree as seen by an observer from a point on the level ground at a distance of 30 m from the foot of the tower is 60^o. Find the height of the tree.

Here, Let AB = height of the tree and C is the position of the observer on the level ground, & BC = 30 m. .....

11. A boy flying a kite paid out a cord of length 150 m. If the angle of elevation of the kite from a point on the ground is 30^o, what will be the height of the kite from the level ground?

Let A be the position of the kite. BA = height of the kite from the level ground CA = length of the cord paid out. Here, CA = 150 m. In right angled triangle ABC, Sin 30o = ABCA or, 12 .....

12. A circular pond has a pole standing vertically at its centre. The top of the pole is 30 m above the water surface and the angle of elevation of it from the point on the circumference is 60^o. Find the radius of the pond.

Let, OP = length of the pole fixed at centre O of circular pond OQ = radius of the pond. Then, OP = 30 m. In right angled triangle OPQ, Tan 60o= OPOQ or, √3 = 30mOQ or, OQ = .....

13. A man 1.5 m tall observes the angle of elevation of the top of the tower as 45^o. If the height of the tower is 50 m, find the distance between the man and the tower.

Let, AB = height of the man CD = height of the tower AE = BD = distance between the man and the tower. Here, AB = 1.5 m CD = 50 m and ∠EAC = 45o In right angled .....

14. A vertical pole is 10√3 m high and its shadow is of length 30 m. What is the angle of elevation of the sun?

Here, Let AC = height of the pole BC = length of the shadow of the pole. Here, AC = 10√3 m and BC = 30 m. In right angled triangle ABC, Tanθ = ACBC Or, tanθ = 10√3m30m Or, tan θ = √33 Or, .....

15. In ∆ABC, if AC = 8 cm, BC = 6 cm and the area of ∆ABC = 24√2cm², find ∡C.

Here, AC = 8 cm and BC = 6 cm Area of ∆ABC = 12√2 cm2 Area of ∆ABC = ½ ⨉ BC ⨉ CA SinC or, 12√2 = ½ ⨉ 6 ⨉ 8 ⨉ SinC .....

16. A man of height 1.5 m observes the angle of elevation of the top of a pole situated in front of him and finds to be 60o. If the height of the pole is 121.5 m, find the distance betweeen the pole and the man.

Here, Let AB to the man and DC be the tower. Distance between them = AM. In triangle AMD, Tan600 = DMAM Or, AM = DMtan60 = 120√3 m .....

17. In the adjoining figure, PS || QR. If PS = 24 cm, PQ = 12 cm, angle SPQ = 60o, angle SQR = 30oand area of PSQ = 2 ⨉ area of QSR, find the length of QR.

Here, Here, ∡ PSQ = 30o [Angles in alternate segment between parallel lines are equal] Also , ∡PQR= 180o–(60 + 30) = 90o. In triangle PQS, using Pythagoras theorem, PS2 = PQ2 + SQ2 Or, SQ = 12√3cm. Area of .....

18. In the adjoining figure, AB = 12 cm, BD = 18 cm, BC = 14 cm, angle CBD = 1/2 angle and angle ABD 54 cm². Find the area of quadrilateral ABCD.

In triangle ABD, Area= 12 ⨉ 12 ⨉ 18 ⨉ sin(2x) Or, 54 = 108 sin (2x). Or, sin2x = 12 Or, sin2x = sin30o Therefore, x = 15o. Now, area of triangle BCD is: Area = 12 ⨉ BC⨉ BD ⨉ sin 15o .....

19. From the top of a tower by the seaside 150 m high, it was observed that the angle of depression of the bottom of a ship was 30^o. Find the distance of the ship from the foot of the tower.

Let AC = height of the tower and B is the position of the ship. BC = distance of the ship from the foot of the tower. In right angled triangle ABC, Tan30o =ACBC .....

20. The upper part of a straight tree is broken by wind and makes an angle of 45^o with the plain surface at point 7m from the foot of a tree. Find the height of the tree before it was broken.

Here, Let A’B be the tree before it was broken. In triangle AMB, Tan45 = 7MB Or, MB = 7m. Again, sin45o = 7mAM Or, AM = 7√2m. Therefore, the height of the original tree = MB +AM .....

21. On a windy day, a boy of height 1.1 m was flying his kite. When the length of the string of the kite was 33m, it makes an angle of 30o with the horizon. At what height was the kite flying?

Here, Let AB denote the man, D be the position of the kite. In triangle DAM, Sin 30o = DM33m Or, 12 = DM33m Therefore, DM = 16.5m. Now, Total height of kite from ground = 16.5m + 1.1m = 17.6 .....

22. A girl is 1.54 m tall, is 30 m away from a tower whose height is 53.5 m. Determine the angle of elevation from her eye to the top of the tower.

Here, Le AB be the tower, and DC be the man. AM = 53.5 – 1.54 = 51.96m. Also, DM = 30 m. So, in triangle ADM, Tan ∡ADM = AMDM Or, Tan ∡ADM = .....

23. A tree of 15 m height is broken by the wind so that its top touches the ground and makes an angle of 60^o with the ground. Find the length of the broken part of the tree.

Here, Let AB be the tree. M be the part from which it is broken. So, In triangle AMB, Cos60o = x15 - x Or, ½ = x15 - x Or, 2x = 15 – x Or, x = 5m. So, 14 –x = 14 - 5 = 9m. Therefore, height of the .....

24. At the center of a circular pond, there is a pole of 11.62 m height above the surface of water. From the point on the edge of the pond, a man of height 1.62m, observed the angle of elevation of top of the pole and found 30o. Find the diameter of the pond.

Here, Let AB denote the man, CD be the position of the tower. In triangle DAM, DM = 11.62 - 1.62 = 10 & tan30o = DMAM Or, √3 = 10mAM Or, AM = 10√3m. Now, Diameter = 2 ⨉ .....

25. In ∆PQR, PR = 120 mm, PQ = 92 mm, ∠P = 29^o, find the area of the ∆PQR given that Sin29^o = 0.4848.

Here, PR = 120 mm PQ = 92 mm ∠P = 29o . We know, Area of △PQR = ½ . PR. PQ SinP .....

26. In ∆XYZ, x = 44 mm, z = 55 mm,∠XYZ = 66^o, find the area of the ∆XYZ given that Sin66^o = 0.914.

Here, x = 44 mm z = 55 mm ∠Y = 66o . We know, Area of △ XYZ = ½ . x.z .SinY .....

27. The thread of a kite makes an angle of 60^o with the horizon while a boy of height 1.8 m is flying a kite. If the length of the thread is 300 m, find the height of the kite from the ground.

Let, AD = height of the boy = 1.8 m AC = length of the thread = 300 m CE = height of the kite from the ground ∠BAC = 60o In right angled triangle .....

28. In ∆ABC, BC = 10 cm, ∡B = 30^o and the area of the ∆ABC = 40 cm², find the length of the side AB.

Here, BC = 10 cm ∡B = 30o Area of ∆ABC = 40 cm2 Area of ∆ABC = ½ . BC .....

29. A man of height 1.6 m stands at a distance of 4.68 m from a lamp-post and it is observed that his shadow is 2.88 m long, find the height of the lamp-post.

Let, AB = height of the man = 1.6 m CD = height of the lamp-post EB = length of the shadow = 2.88 m BD = distance of the man from the lamp-post. In right angled .....

30. A man 1.6 m tall has the shadow of length 1.2 m. Find the length of the shadow of a tree 12.40 m tall at the same time of the day.

Let, BC = height of the man = 1.6 m AB = shadow of the man = 1.2 m Height of the tree (H) = 12.40 m Shadow of the tree = x ∠CAB = θ = .....

31. In ∆ABC, a = 13 cm, b = 10 cm, ∠C = 65^o, find the area of the ∆ABC given that Sin65^o = 0.906.

Here, a = 13 cm b = 10 cm ∠C = 65o . We know, Area of△ABC = ½ . ab SinC .....

32. An observer of height 1.6 m is standing 75 m away from the foot of a tower. He observes the top of the tower and finds the angle of elevation to be 30^o. Find the height of the tower.

Let, AD = height of the man = 1.6 m CE = height of the tower AB = DE = distance between the man and the tower = 75 m ∠BAC = 30o In right angled triangle .....

33. In ∆ABC, b = 3.5 cm, a = 7.1 cm, ∠C = 21.5^o, find the area of the ∆ABC given that Sin21.5^o = 0.367.

Here, b = 3.5 cm a = 7.1 cm ∠C = 100o . We know, Area of △ABC = ½ . ab SinC .....

34. In ∆ABC, if CA = 6 cm, AB = 5 cm and area of the ∆ABC = 7.5 cm², find ∠A.

Given, CA = b = 6 cm AB = c = 5 cm Area of ∆ABC = 7.5 cm2 We know, Area of ∆ABC = .....

35. A man of height 1.5 m, standing 150m away from a pole, found the angle of elevation to be 45o while observing the pole. Find the height of the pole.

Here, Let AB denote the man, DC the tower and BC the distance between the tower and man. In triangle ADM, Tan ∡DAM = DMAM Or, tan45o = DM150m [Since, .....

36. When the sun’s altitude is 60^o, the length of the shadow of a vertical pole is 6√3 m, find the height of the pole.

Let, BC = height of the pole AB = length of the shadow = 6√3 m. In right angled triangle ABC, Tan60o = BCAB or, √3 = BC6√3m or, BC = 6√3m ⨉ √3 or, BC = 18 m. So, the .....

37. A tree on the bank of a river is 20√3 m high and the angle of elevation of the top of the tree from the opposite bank is 30^o. Find the breadth of the river.

Here, Let AC = height of the tree And BC = breadth of the river. In right angled triangle ABC, Tan 30o = ACBC or, 1√3 = 20√3BCm or, BC = 20√3m ⨉ √3 or, .....

38. In ∆ABC, a = 4.5 cm, c = 3.9 cm, ∠B = 46^o, find the area of the ∆ABC given that Sin46^o = 0.72.

Here, a = 4.5 cm c = 3.9 cm ∠B = 46o Area of triangle ABC = ½ . ac SinB .....

39. In ∆ABC, if AC = 8 cm, BC = 5 cm and the area of ∆ABC = 10√2cm², find ∡C.

Here, AC = 8 cm and BC = 5 cm. Area of ∆ABC = 10√2 cm2. We know, Area of ∆ABC = ½ ⨉ BC ⨉ CA SinC or, 10√2 = ½ ⨉ 5 ⨉ 8 ⨉ SinC or, 10√2 .....

40. In ∆ABC, if BC = 10 cm, ∠C = 60^o and area of the ∆ABC = 25 cm², find the length of AC.

Given, BC = 10 ∡C = 60o Area of ∆ABC = 25 cm2 We know, Area of ∆ABC = ½ BC. AC. SinC .....

41. In ∆ABC, if BC = 18 cm, ∡C = 30^o, and the area of ∆ABC = 25 cm², find the length of the side AC.

Here, BC = 18 cm ∡C = 30o Area of ∆ABC = 25 cm2 We know, Area of ∆ABC = ½ ⨉ BC ⨉ AC SinC or, 25cm2 .....

42. In ∆ABC, if BC = 8 cm, ∡C = 30^o and area of the ∆ABC = 12 cm², find the length of AC.

Given, BC = 8 ∡C = 30o Area of ∆ABC = 12 cm2 We know, Area of ∆ABC = ½ BC. AC. SinC or, 12 = ½ ⨉ 8 ⨉ AC Sin30o .....

43. In ∆ABC, if a = 8 cm, b = 5√2 cm and ∡C = 30^ofind the area of the ∆ ABC.

Given, a = 8 cm b = 5√2 cm. Area of ∆ABC = ½ ab SinC = ½ 8 .....

44. A tree 15 m high is broken by the wind such that the top of the broken part touches the ground and makes an angle of 30^o with the ground. Find the length of the broken part of the tree.

Let, AB = Original height of the tree AC = length of the broken part of the tree Let D be the point on the ground at which the top of the tree meets after broken. AC = CD Here, ∠AB = 15 m and ∠BDC = 30o In right angled triangle .....

45. Find the area of the adjoining triangle.

Here, PQ = 8cm, QR = 12cm. The area of the triangle PQR is given by, Area = 12 ⨉PQ⨉QR⨉sin(∡PQR) = 12 ⨉8cm⨉12cm⨉sin30° = 32 .....

46. In the given triangle DEF; DF=7cm, EF=10cm, area of ∆DEF= 3523 cm² and ∡DEF=30°, find the value of ∡EDF.

We have, or, 12 ⨉ DE ⨉ EF ⨉ sin(∡DEF) = 12⨉ DE⨉ DF ⨉ sin(∡EDF) [Both are areas of same triangle] or, 5 ⨉sin(30°) = 7 ⨉ sin(∡EDF) or, 57 = sin(∡EDF) or,∡EDF = sin-1 (57) = .....

47. Find the area of the given kite.

Here, ∡ ADB = ∡ BDC [ Diagonal of kite bisect the angle] Now, Area of ∆ABD = 12⨉AD⨉BD⨉sin(∡ADB) = 12 ⨉ 3⨉5⨉sin(30°) .....

48. In the given figure CD=1.5 m, CB=150 m and ∡ADE=45°, then find the measure of AB.

Here, DE = CB = 150 m [ Opposite sides of rectangle CDEB] In right angled ∆ADE, Tan45°=AEDE Or, 1 = AEDE ∴AE = DE = 150 m AB = AE + EB = AE + EB .....

49. In the given figure, AB = 12 cm, BC = 16√3 cm, AD = 18 cm, angle ABC = 60^o and angle ADC = 30^o If area of triangle ACD = 14^th ABC, find the length of CD.

Here, We know, Area of triangle ABC = 12 ⨉ AB ⨉ BC⨉ sin60 = .....

50. In ∆ABC, BC = 8 cm, ∡B = 30^o and the area of the ∆ABC = 20 cm², find the length of the side AB.

Solution: Here, BC = 8 cm ∡B = 30o Area of ∆ABC = 20 cm2 Area of ∆ABC = ½ . BC . AB SinB or, 20 = ½ ⨉ 8 ⨉ AB ⨉ Sin30o or, .....