10类RD Sharma解决方案–第5章三角比–练习5.2 |套装3

问题27.如果A = B = 60°，请确认

(i)cos(AB)= cos A cos B + sin A sin B

解决方案：

We know sin 60° = √3/2, sin 30° = 1/2

Putting the values of A and B in the equation below

cos(60° − 60°) = cos60°cos60° + sin60°sin60°

cos(0) = cos²60° + sin²60°

1 = (1/2)² + (√3/2)²

1 = 1/4 + 3/4

1 = 4/4 = 1 

Hence Proved

(ii)sin(AB)= sin A cos B − cos A sin B

解决方案：

Putting the values of A and B in the equation below

sin(60° − 60°) = sin60°cos60° − cos60°sin60°

sin(0°) = (1/2)(√3/2) – (√3/2)(1/2) 

0 = √3/4 – √3/4

0 = 0  

Hence Proved

(iii)tan(A-B)= $\frac{tanA−tanB}{1+tanAtanB}$

解决方案：

Putting the values of A and B in the equation below

tan(60° − 60°) = $\frac{tan60°−tan60°}{1+tan60°tan60°}$

tan0 =$\frac{0}{1+tan60°tan60°}$

0 = 0 

Hence Proved

问题28.如果A = 30°且B = 60°，请验证

(i)sin(A + B)= sin A cos B cos A sin B

解决方案：

We know sin 60° = cos 30° = √3/2, sin 30° = cos 60° = 1/2

Putting the values of A and B in the following equations

sin(30° + 60°) = sin 30° cos 60° cos 30° sin 60°

sin 90° = (1/2)(1/2) + (√3/2)(√3/2)

1 = 1/4 + 3/4

1 = 1 

Hence Proved

(ii)cos(A + B)= cos A cos B − sin A sin B

解决方案：

Putting the values of A and B in the following equations

cos(30° + 60°) = cos 30° cos 60° – sin 30° sin 60°

cos 90° = (√3/2)(1/2) – (1/2)(√3/2)

0 = √3/4 – √3/4

0 = 0 

Hence Proved

问题29.如果sin(A + B)= 1且cos(A-B)= 1，则0°

解决方案：

Given sin(A + B) = 1, cos(A − B) = 1, 0° < A + B ≤ 90° and A ≥ B,

We know that, sin90° = 1 and cos0° = 1

sin(A + B) = 1 and sin90∘= 1

We can say that

(A + B) = 90°          -(1)

cos(A − B) = 1 and cos0° = 1

(A − B) = 0°

A = B          -(2)   

Substituting eq(2) in (1)

A + A = 90°

2A = 90°

A = 45°

A = B = 45°

问题30.如果tan(A + B)=√3且tan(A-B)= 1 /√3，则0° B，找到A和B。

解决方案：

Given tan(A + B) = √3

tan(A − B) = 1/√3

0∘< A + B ≤ 90∘and A > B,

We know that, tan30° = 1/√3 and tan60° = √3 

Now,

tan(A + B) = √3 and tan60 = √3

(A + B) = 60°          -(1)

Again,

tan(A − B) = 1/√3 and tan30° = 1/√3

(A − B) = 30°          -(2)

Solving eq(1) and (2),

(A + B) = 60°

(A − B) = 30°

Solving the above two equations we get

2A = 90°

A = 90°/2 = 45°

问题31.如果sin(A – B)= 1/2且cos(A + B)= 1/2，则0°

解决方案：

Given, sin(A − B) = 1/2

cos(A + B) = 1/2

0° < A + B ≤ 90° and A < B,

We know that, sin30° = 1/2and cos60° = 1/2

Since sin(A − B) = 1/2 and sin30° = 1/2

So (A − B) = 30°          -(1)

Again  

cos(A + B) = 1/2 and cos60° = 1/2

So (A + B) = 60°          -(2)

Solving eq(1) and (2) we get

2A = 90°

A = 90°/2 = 45°

From (2)

(A + B) = 60°

Also, A = 45°

B = 60° − 45°

B = 15°

Therefore, A = 45°, B = 15°

问题32.在与B成直角的∆ABC中，∠A=∠C。查找的值

(i)罪A cos C +罪A A sin C

解决方案：

Given: In ∆ABC right-angled at B, ∠A = ∠C

We know in a triangle ∠A + ∠B +∠C = 180°

2∠A = 180° – 90° = 45°               [∠B = 90°]

Now,

We have sin A cos C + cos A sin C

Putting the values of each angle 

sin 45° cos 45° + cos 45° sin 45° 

= (1/√2)(1/√2) + (1/√2)(1/√2)

= 1/2 + 1/2

= 1

(ii)罪A罪B + cos A cos B

解决方案：

We have sin A sin B + cos A cos B

Putting the values of each angle  

sin 45° sin 90° + cos 45° cos 90°  

= (1/√2)(1) + (1/√2)(0)

= (1/√2) + 0

= 1/√2

问题33.如果sin(A + 2B)=√3/ 2且cos(A + 4B)= 0，并且A> B，则求出锐角A和B。

解决方案：

Given, sin(A + 2B) = √3/2, cos(A + 4B) = 0, A > B,

We know that, sin60° = √3/2 and cos90° = 0

Since sin(A + 2B) = √3/2 and sin60° = √3/2

So (A + 2B) = 60°          -(1)

Again cos(A + 4B) = 0 and cos90° = 0

So (A + 4B) = 90°          -(2)

From (1) from (2) we get,

2B = 30°

B = 30°/2 = 15°

From (2)

(A + 4B) = 90°

Also, B = 15°

A = 90° − 60°

A = 30°

Therefore, A = 30°, B = 15°

问题34.在ΔPQR中，在Q处成直角，PQ = 3 cm，PR = 6 cm。确定∠P和∠R。

解决方案：

Given In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm.

We know cos P = Base/Hypotenuse =3/6

cos 60° = 1/2

So ∠P = 60°

In ΔPQR 

∠P + ∠Q + ∠R = 180°

∠R = 180° – 90° – 60°

∠R = 30°

Therefore, ∠P = 60°, ∠R = 30°

问题35.如果sin(A-B)= sin A cos B-cos A sin B和cos(A-B)= cos A cos B + sin A sin B，则求出sin15°和cos15°的值。

解决方案：

To find sin15°:

We have sin15° = sin(45° – 30°)

A = 45°and B = 30°

Now putting the values of A and B in the following equation

sin(A − B) = sin A cos B − cos A sin B

sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30°

sin15° = (1/√2)(√3/2) – (1/√2)(1/2))

sin15° = (√3 – 1)/2√2 

To find cos15°:

 cos15° = cos(45° – 30°)

A = 45° and B = 30°

Again putting the values of A and B in the following equation

 cos(A − B) = cos A cos B + sin A sin B

cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30°

sin15° = (1/√2)(√3/2) + (1/√2)(1/2))

sin15° = (√3 + 1)/2√2   

问题36.如果∠B= 60°且AB = 15单位，则在右△ABC中，与C成直角。找到剩余的角度和边。

解决方案：

Given In a right △ABC right-angled at C, ∠B = 60° ∠C = 90°

We know in a triangle ∠A + ∠B + ∠C = 180°

∠A = 180° – 150° = 30°

sin 30° = BC/AB = BC/15 

BC/15 = 1/2

BC = 7.5 units

cos 30° = AC/AB = AC/15

AC/15 = √3/2

AC = 15√3/2 units

问题37.如果ΔABC是直角三角形，使得∠C= 90°，∠A= 45°，并且BC = 7个单位。查找∠B，AB和AC。

解决方案：

Given: In ΔABC ∠C = 90°, ∠A = 45°, and BC = 7 units

We know in a triangle ∠A + ∠B + ∠C = 180°

∠B = 180° – 135° = 45°

cos B = BC/AB = 7/AB

cos 45° = 7/AB

7/AB = 1/√2

AB = 7/√2 units

tan A = BC/AC

tan 45° = 7/AC

7/AC = 1

AC = 7 units

问题38.在矩形ABCD中，AB = 20 cm ，, BAC = 60°，计算边BC和对角线AC，BD。

解决方案：

Given: rectangle ABCD, AB = 20 cm, ∠BAC = 60°

tan 60° = BC/AC

√3 = BC/20

BC = 20√3 cm

cos 60° = AB/AC

1/2 = 20/AC

AC = 40 cm

In rectangle both diagonals are equal

问题39.如果A和B是锐角，使得tan A = 1/2，tan B = 1/3，并且tan(A + B)= $\frac{tanA+tanB}{1−tanAtanB}$找到(A + B)。

解决方案：

Given, A and B are acute angles such that tanA = 1/2 and tanB = 1/3

tan(A + B) = $\frac{tanA+tanB}{1−tanAtanB}$

tan(A + B) = $\frac{\frac{1}{2}+\frac{1}{3}}{1−(\frac{1}{2})(\frac{1}{3})}$

tan(A + B) = $\frac{\frac{5}{6}}{1−\frac{1}{6}}$

tan(A + B) = $\frac{\frac{5}{6}}{\frac{5}{6}}$

tan(A + B) = 1

We know, tan45° = 1

Therefore, (A + B) = 45°

问题40.证明(√3+1)(3 – cot 30°)= tan ³ 60°– 2sin 60°

解决方案：

Taking R.H.S

tan³ 60° – 2sin 60°

(√3) – 2(√3/2) = 3√3 – √3 = 2√3

Now taking L.H.S

(√3 + 1)(3 – cot 30°) 

= (√3 + 1)(3 – √3) 

= (√3 + 1)√3(√3 – 1) 

= 2√3

L.H.S = R.H.S

Hence Proved