10类RD Sharma解决方案–第5章三角比–练习5.1 |套装1

问题1.在下面的每一个中，给出六个三角比之一。查找其他三角比例的值。

(i)sinA = 2/3

解决方案：

sinA = 2/3 = Perpendicular/Hypotenuse                                                        

Draw a right-angled △ABC in which ∠B is = 90°                       

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(3)² = (2)² + (BC)² 

 9 = 4 + BC² 

 BC² = 9 – 4 = 5

BC = √5 units         

Now,

cosA = Base/Hypotenuse = BC/AC = √5/3   

tanA = Perpendicular/Base = AB/BC = 2/√5         

cotA = 1/tanA = √5/2          

secA = 1/cosA = 3/√5         

cosecA = 1/sinA = 3/2          

(ii)cosA = 4/5

解决方案：

cosA = 4/5 = Base/Hyptenuse

Draw a right-angled △ABC in which ∠B is = 90° 

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(5)² = (AB)² + (4)²

 25 = AB² + 16 

AB² = 25 – 16 = 9

AB = √9

= 3 units     

Now, 

sinA = Perpendicular/Hypotenuse = AB/AC =3/5        

tanA = Perpendicular/Base = AB/BC = 3/4        

cotA = 1/tanA = 4/3

secA = 1/cosA = 5/4

cosecA = 1/sinA =5/3

(iii)tanθ= 11/1

解决方案：

tanθ = 11/1 = Perpendicular/Base

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

AC² = (11)² + (1)²

AC² = 121  + 1

= 122

AC = √122units  

 Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 11/ √122

cosθ = Base/Hypotenuse = BC/AC = 1/√122

cotθ = 1/tanθ = 1/11         

secθ = 1/cosθ = √122/1

cosecθ = 1/sinθ = √122/11     

(iv)sinθ= 11/15

解决方案：

sinθ = 11/15 = Perpendicular/Hypotenuse

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(15)² = (11)² + (BC)²

225 = 121 + (BC)²

(BC)² = 104

BC = 2√26

Now,

cosθ = Base/Hypotenuse = BC/AC = 2√26/15              

tanθ = AB/BC = 11/ 2√26      

cotθ = 1/tanθ = 2√26/11    

secθ = 1/cosθ = 15/ 2√26     

cosecθ = 1/sinθ = 15/11             

(v)棕褐色α= 5/12

解决方案：

tan α = 5/12 = Perpendicular/Base

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(AC)² = (12)² + (25)²

(AC)² = 144 + 25 

(AC)² = 169

AC = √169 = 13 units  

Now,   

sin α = Perpendicular/Hypotenuse = AB/AC = 5/13     

cos α = Base/Hypotenuse = BC/AC = 12/13           

cot α = 1/tan α = 12/5     

sec α = 1/cos α = 13/12      

cosec α = 1/sin α = 13/5  

(vi)sinθ=√3/ 2

解决方案：

sinθ = √3/2 = Perpendicular/Hypotenuse 

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(2)² = (√3​)² + (BC)²

4 = 3 + (BC)²

(BC)² = 4 – 3 = 1 

BC = 1 units

Now,

cosθ = Base/Hypotenuse = BC/AC = 1/2         

tanθ = AB/BC = √3/1      

cotθ = 1/tanθ = 1/√3

secθ = 1/cosθ = 2/1    

cosecθ = 1/sinθ = 2/√3        

(vii)cosθ= 7/25

解决方案：

 cosθ = 7/25 = Base/Hypotenuse  

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(25)² = (AB)² + (7)²

625 = (AB)² + 49

(AB)² = 625 – 49 = 576

AB = √576 = 24 units

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 24/25         

tanθ = Perpendicular/Base = AB/BC = 24/7     

cotθ = 1/tanθ = 7/24         

secθ = 1/cosθ = 25/7        

cosecθ = 1/sinθ = 25/24                   

(viii)tanθ= 8/15

解决方案：

tanθ = 8/15 = Perpendicular/Base  

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(AC)² = (8)² + (15)²

(AC)² = 64 + 225 

AC = √289 = 17

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 8/17       

cosθ = Base/Hypotenuse = BC/AC = 15/17       

cotθ = 1/tanθ = 15/8       

secθ = 1/cosθ = 17/15      

cosecθ = 1/sinθ = 17/8      

(ix)cotθ= 12/5

解决方案：

cotθ = 12/5 = Base/Perpendicular     

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(AC)² = (5)² + (12)²

(AC)² = 25 + 144 

(AC)² = 169

AC = √169 = 13 units 

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 5/13     

cosθ = Base/Hypotenuse = BC/AC = 12/13     

tanθ = 1/tanθ = 5/12     

secθ = 1/cosθ = 13/12    

cosecθ = 1/sinθ = 13/5   

[x)秒θ= 13/5

解决方案：

secθ = 13/5 = Hypotenuse/Base 

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(13)² = (AB)² + (5)²

169 = (AB)² + 25

(AB)² = 169 – 25 = 144

AB = √144 = 12 units

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 12/13      

tanθ = Perpendicular/Base = AB/BC = 12/5    

cotθ = 1/tanθ = 5/12     

cosθ = 1/secθ = 5/13    

cosecθ = 1/sinθ = 13/12     

(xi)cosecθ=√10

解决方案：

cosecθ = √10/1 = Hypotenuse/Perpendicular 

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(√10)² = (1)² + (BC)²

10 = 1 + (BC)²

(BC)² = 10 – 1 = 9

BC = √9 = 3

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 1/√10     

cosθ = Base/Hypotenuse = BC/AC = 3/√10   

tanθ = Perpendicular/Hypotenuse = AB/BC = 1/3   

cotθ = 1/tanθ = 3/1 = 3           

secθ = 1/cosθ = √10/3        

(xii)cosθ= 12/15

解决方案：

cosθ = 12/15 = Base/Hypotenuse

Draw a right-angled △ABC in which ∠B is = 90°

Using Pythagoras Theorem, in △ABC,

(Hypotenuse)² = (Perpendicular)² + (Base)²

AC² = AB² + BC²

(15)² = (AB)² + (12)²

225 = (AB)² + 144

(AB)² = 225 – 144 = 81

AB = √81 = 9 units   

Now,

sinθ = Perpendicular/Hypotenuse = AB/AC = 9/15

tanθ = Perpendicular/Base = AB/BC = 9/12    

cotθ = 1/tanθ = 12/9    

secθ = 1/cosθ = 15/12      

cosecθ = 1/sinθ = 15/9      

问题2.在ΔABC中，与B成直角，AB = 24 cm，BC = 7 cm。决定

(i)罪A，cos A

(ii)罪C，cos C

解决方案：

Given:

In right-angled ΔABC,

AB = 24 cm, BC = 7 cm. ∠B = 90°

Using Pythagoras Theorem 

AC² = AB² + BC²

AC² = 24² + 7² = 576 + 49

AC² = 625

AC = √625 = 25cm

Now,

(i) sinA = BC/AC = 7/25

cosA = AB/AC = 24/25    

(ii) sinC = AB/AC = 24/25 

cosC = BC/AC = 7/25    

问题3.在图中，找到tan P和cotR。tanP = cot R?

解决方案：

Using Pythagoras Theorem

PR² = PQ² + QR²

13² = 12² + QR²

QR² = 169 – 144 = 25

QR = √25 = 5 cm

Now,

tan P = Perpendicular/Base = QR/PQ = 5/2  

cot R = Base/Perpendicular = QR/PQ = 5/2  

Yes, tanP = cot R

问题4.如果sin A = 9/41，则计算cos A和tanA。

解决方案：

Given, sinA = 9/41 = Perpendicular/Hypotenuse

Draw a △ ABC where ∠B = 90°, BC = 9, AC = 41 

Using Pythagoras Theorem

AC² = AB² + BC²

BC² = 41² – 9² = 1681 – 81

BC² = 1600

BC = √1600 = 40

Now, cos A = Base/Hypotenuse = AB/AC = 40/41 

tan A = Perpendicular/Base = BC/AB = 9/40                   

问题5.给定15婴儿床A = 8，求正A和secA。

解决方案：

Given, 15 cot A = 8 

cot A = 8/15 = Base/Perpendicular

Draw a △ ABC where ∠B = 90°, AB = 8, BC = 15

Using Pythagoras Theorem

AC² = AB² + BC²

AC² = 8² + 15² = 64 + 225

AC² = 289

AC = √289 = 17

Now,

sin A = Perpendicular/Hypotenuse = BC/AC = 15/17      

sec A = Hypotenuse/Base = AC/AB = 17/8                

问题6.在ΔPQR中，在Q处成直角，PQ = 4 cm，RQ = 3 cm。求出sin P，sin R，sec P和sec R的值。

解决方案：

In right-angled ΔPQR,

∠Q = 90°, PQ = 4cm, RQ = 3cm

Using Pythagoras Theorem

PR² = PQ² + QR²

PR² = 4² + 3² = 16 + 9

PR² = 25

PR = √25 =5 

Now,

sin P = Perpendicular/Hypotenuse = RQ/PR = 3/5          

sin R = Perpendicular/Hypotenuse = PQ/PR = 4/5         

sec P = Hypotenuse/Base = PR/PQ = 5/4    

sec R = Hypotenuse/Base = PR/RQ = 5/3