## Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry## Exercise 9.1
It can be observed that ABC is a right angled triangle with AC as the hypotenuse. Therefore, sin C = AB/AC sin 30° = AB/20 1/2 = AB/20 AB = 20/2 = 10 m Hence, the height of the vertical pole is 10 m.
Let the remaining vertical part of tree be AB and the fallen part be AC. It can be noted that ABC forms a right angled triangle with AC as its hypotenuse. Thus, cos C = BC/AC cos 30° = 8/AC √3/2 = 8/AC AC = 16/√3 m tan C = AB/BC 1/√3 = AB/8 AB = 8/√3 m Height of the tree = AB + AC = 8/√3 m + 16/√3 m = 24/√3 m Rationalise, 24/√3 × √3/√3 = 8√3 m Hence, the height of the tree is 8√3 m.
Let the slide for children younger than 5 years be represented by ABC and the slide for elder children be represented by PQR. AB = 1.5 m and PQ = 3 m. In ABC, we have sin C = AB/AC sin 30° = 1.5/AC 1/2 = 1.5/AC AC = 3 m In PQR, we have sin R = PQ/PR sin 60° = 3/PR √3/2 = 3/PR PR = 6/√3 m Rationalise, 6/√3 × √3/√3 = 2√3 m Hence, the length of slide for younger children = 3 m, and length of slide for elder children = 2√3 m
Let AB be the tower and C be the point 30 m away from the foot B. tan C = AB/BC tan 30° = AB/30 1/√3 = AB/30 AB = 30/√3 m Rationalize, 30/√3 × √3/√3 = 10√3 m Hence, the height of the tower is 10√3 m.
Let PQ be the height of the kite from ground and PR be the string attached to the kite. sin R = PQ/PR sin 60° = 60/PR √3/2 = 60/PR PR = 120/√3 m Rationalise, 120/√3 × √3/√3 = 40√3 m Hence, the length of the string assuming it has no slack is 40√3 m.
Let the above figure represent the given situation. AE = Height of Building = 30 m CF = Height of the boy = 1.5 m BC = Distance between boy and the building DC = Displacement of the boy from his original position C In ABC, we have tan C = AB/BC tan 30° = (AE - CF)/BC 1/√3 = (30 - 1.5)/BC BC = 28.5√3 m In ABD, we have tan D = AB/BD tan 60° = 28.5/BD √3 = 28.5/BD BD = 28.5/√3 m Rationalize, 28.5/√3 × √3/√3 = 9.5√3 m CD = BC - BD = 28.5√3 m - 9.5√3 m = 19√3 m Hence, the boy walked 19√3 m towards the building.
Let AB be the transmission tower on top of building BC. The required point on the ground is E. In BCE, we have tan CEB = BC/CE tan 45° = 20/CE 1 = 20/CE CE = 20 m In ACE, we have tan CEA = AC/CE tan 60° = AC/20 √3 = AC/20 AC = 20√3 m Height of transmission tower = AB = AC - BC = 20√3 - 20 = 20(√3 - 1) m Hence, height of the transmission tower = 20(√3 - 1) m
Let AB be the statue on the pedestal BC and let E be the point on the ground. In ACE, we have tan CEA = AC/CE tan 60° = (AB + BC)/CE √3 = (1.6 + BC)/CE (1.6 + BC)/√3 = CE In BCE, we have tan CEB = BC/CE tan 45° = BC/CE 1 = BC/CE CE = BC Therefore, (1.6 + BC)/√3 = BC 1.6 + BC = BC√3 1.6 = BC(√3 - 1) BC = 1.6/(√3 - 1) Rationlaize, BC = 1.6/(√3 - 1) × (√3 + 1)/(√3 + 1) BC = 1.6(√3 + 1)/2 BC = 0.8(√3 + 1) m Hence, height of the pedestal is 0.8(√3 + 1) m.
Let AB be the building and DC be the tower. In BCD, we have tan B = DC/BC tan 60° = 50/BC √3 = 50/BC BC = 50/√3 Rationalise, BC = 50/√3 × √3/√3 = 50√3/3 m In ABC, we have tan C = AB/BC 1/√3 = AB/(50√3/3) AB = 50/3 m Hence, height of the building is 50/3 m.
Let AB and DC be the poles on road BC and let the required point on the road be E. In DCE, we have tan E = CD/EC tan 30° = CD/EC 1/√3 = CD/EC CD = EC/√3 In ABE, we have tan E = AB/BE tan 60° = AB/BE √3 = AB/BE AB = BE√3 We know that AB = CD (Given). Therefore, EC/√3 = BE√3 We also know that EC = 80 - BE. Therefore, (80 - BE)/√3 = BE√3 80 - BE = 3BE 80 = 4BE BE = 20 m Therefore, EC = 80 - 20 = 60 m AB = CD = 20√3 m Hence, the height of the poles is 20√3 m and the distance of the point from the poles is 20 m and 60 m respectively.
In ABD, we have tan D = AB/BD tan 30° = AB/BD 1/√3 = AB/(20 + BC) AB = (20+BC)/√3 In ABC, we have tan C = AB/BC tan 60° = AB/BC √3 = AB/BC AB = √3 BC Therefore, √3 BC = (20 + BC)/√3 3BC = 20 + BC 2 BC = 20 BC = 10 m AB = 10√3 m Hence, height of the tower is 10√3 m and width of the canal is 10 m.
Let AB be the 7 m high building and let the cable tower be EC. AD is an imaginary line drawn parallel to the ground BC. AB = CD = 7 m In ACD, we have tan A = CD/AD tan 45° = 7/AD 1 = 7/AD AD = 7 m In ADE, we have tan A = DE/AD tan 60° = DE/7 √3 = DE/7 DE = 7√3 m EC = DE + CD = 7√3 + 7 = 7(√3 + 1) m Hence, the height of the cable tower is 7(√3 + 1) m.
Let AB be the lighthouse and C and D be the positions of the ships in the sea respectively. Draw an imaginary line AE which is parallel to the sea level BD. In ABC, we have tan C = AB/BC tan 45° = AB/BC 1= 75/BC BC = 75 m In ABD, we have tan D = AB/BD tan 30° = AB/BD 1/√3 = 75/BD BD = 75√3 m CD = BD - BC = 75√3 - 75 = 75(√3 - 1) m Hence, distance between the two ships in the sea = 75(√3 - 1) m
Let A and B be the initial and final positions of the balloon. Let C be the head of the girl. AD and BE are imaginary lines drawn perpendicular to the ground from the positions of the balloon. AD = BE = 88.2 - 1.2 = 87 m (Unchanged elevation in the balloon) In ACD, we have tan C = AD/CD tan 60° = AD/CD √3 = 87/CD CD = 87/√3 Rationalize, CD = 87/√3 × √3/√3 = 29√3 m In BCE, we have tan C = BE/CE tan 30°= 87/CE 1/√3 = 87/CE CE = 87√3 m DE = CE - CD = 87√3 - 29√3 = 58√3 m Hence, the distance travelled by the balloon is 58√3 m.
Let AB be the tower. Let D and C be the car's initial and final positions respectively. Draw an imaginary line AE which is parallel to the ground. In ABC, we have tan C = AB/BC tan 60° = AB/BC √3 = AB/BC BC = AB/√3 AB = √3 BC In ABD, we have tan D = AB/AD tan 30° = AB/BD 1/√3 = AB/BD AB = BD/√3 Rationalize, AB = BD/√3 × √3/√3 = BD√3/3 Therefore, BD√3/3 = √3 BC BD = 3BC 3BC = BC + CD 2BC = CD BC = CD/2 Time taken to travel distance CD = 6 s Hence, time taken to travel distance BC = CD/2 = 6/2 = 3 s
Let AB be the tower and C and D be the two required points on the ground. Let angle ACB be x, then angle ABD = 90 - x In ABC, we have tan x = AB/BC tan x = AB/4
In ABD, we have tan (90°-x) = AB/BD cot x = AB/9
Multiply both the obtained equations: AB AB AB AB = 6 m as height cannot be negative. Hence, the height of the tower has been proven to be 6 m. Next TopicClass 10 Maths Chapter 10 |