Class 11 RD Sharma解决方案–第10章正弦和余弦公式及其应用–练习10.2 |套装1

问题1.在∆ABC中，如果a = 5，b = 6且C = 60 ^o ，则表明其面积为(15√3)/ 2平方单位。

解决方案：

We are given a = 5, b = 6 and C = 60^o. 

Now we know the area of a ∆ABC is given by 1/2 ab sin C where, a and b are the lengths of the sides of a triangle and C is the angle between these sides.

So, area of ∆ABC = (1/2) 5(6) sin 60^o 

= (1/2) (5) (6) (√3/2)

= 15√3/2 sq. units

Hence, proved.

问题2.在∆ABC中，如果a =√2，则b =√3和c =√5表示其面积为√6/ 2平方单位。

解决方案：

We are given a = √2, b = √3 and c = √5. 

According to Cosine formula, cos C = (a² + b² – c²)/2ab

= (2 + 3 – 5)/2√6

= 0

So, sin C = √(1–cos² C) = 1

Now we know the area of a ∆ABC is given by 1/2 ab sin C where, a and b are the lengths of the sides of a triangle and C is the angle between these sides.

Therefore, area of ∆ABC = (1/2) (√2) (√3) (1)

= √6/2 sq. units.

Hence, proved.

问题3.三角形的边为a = 4，b = 6和c = 8，表明：

8个cos A + 16个cos B + 4个cos C = 17

解决方案：

We are given a = 4, b = 6 and c = 8.

According to Cosine formula, cos A = (b² + c² – a²)/2bc

= (36 + 64 – 16)/96

= 84/96

= 7/8

Also, cos B = (a² + c² – b²)/2ac

= (16 + 64 – 36)/64

= 44/64

= 11/16

Also, cos C = (a² + b² – c²)/2ab

= (16 + 36 – 64)/48

= –12/48

= –1/4

Here, L.H.S. = 8 cos A + 16 cos B + 4 cos C 

= 8 × (7/8) + 16 × (11/16) + 4 × (–1/4) 

= 7 + 11 – 1

= 17

= R.H.S.

Hence, proved.

问题4.在ΔABC中，如果a = 18，b = 24，c = 30，则求出cos A，cos B和cosC。

解决方案：

We are given a = 18, b = 24 and c = 30.  

According to Cosine formula, cos A = (b² + c² – a²)/2bc

= (24² + 30² – 18²)/2(24)(30)

= (576 + 900 – 324)/1440

= 1152/1440

= 4/5

Also, cos B = (a² + c² – b²)/2ac

= (324 + 900 – 576)/2(18)(30)

= 648/1080

= 3/5

Also, cos C = (a² + b² – c²)/2ab

= (324 + 576 – 900)/2(18)(24)

= 0

Therefore, the values of cos A, cos B and cos C are 4/5, 3/5 and 0 respectively.

问题5.对于任何ΔABC ，请证明b(c cos A – a cos C)= c ² – a ²

解决方案：

According to Cosine formula, 

cos A = (b² + c² – a²)/2bc

=> bc cos A = (b² + c² – a²)/2 . . . . (1)

Also, cos C = (a² + b² – c²)/2ab

=> ab cos C = (a² + b² – c²)/2 . . . . (2)

Subtracting (2) from (1), we get,

L.H.S. = bc cos A – ab cos C = (b² + c² – a²)/2 – (a² + b² – c²)/2

= (b² + c² – a² – a² – b² + c²)/2

= (2c² – 2a²)/2

= c² – a²

= R.H.S.

Hence, proved.

问题6.对于任何ΔABC，证明c(a cos B – b cos A)= a ² – b ²

解决方案：

According to Cosine formula,

cos B = (a² + c² – b²)/2ac

=> ac cos B = (a² + c² – b²)/2  . . . . (1)

Also cos A = (b² + c² – a²)/2bc

=> bc cos A = (b² + c² – a²)/2  . . . . (2)

Subtracting (2) from (1), we get,

L.H.S. = ac cos B – bc cos A = (a² + c² – b²)/2 – (b² + c² – a²)/2

= (a² + c² – b² – b² – c² + a²)/2  

= (2a² – 2b²)/2

= a² – b²

= R.H.S.

Hence, proved.

问题7.对于任何ΔABC证明，2(bc cos A + ca cos B + ab cos C)= a ² + b ² + c ²

解决方案：

According to Cosine formula,

cos A = (b² + c² – a²)/2bc

=> 2bc cos A = b² + c² – a²  . . . . (1)

Also, cos B = (a² + c² – b²)/2ac

=> 2ac cos B = a² + c² – b² . . . . (2)

Also, cos C = (a² + b² – c²)/2ab

=> 2ab cos C = a² + b² – c² . . . . (3)

Adding (1), (2) and (3), we get,

L.H.S. = 2bc cos A + 2ac cos B + 2ab cos C 

= b² + c² – a² + a² + c² – b² + a² + b² – c²

= a² + b² + c²

= R.H.S.

Hence, proved.

问题8.对于任何ΔABC，表明(c ² + b ² – a ² )tan A =(a ² + c ² – b ² )tan B =(a ² + b ² – c ² )tan C

解决方案：

According to sine rule in ΔABC,

sin A/a = sin B/b = sin C/c = k (constant) 

According to Cosine formula,

cos A = (b² + c² – a²)/2bc

2bc cos A = (b² + c² – a²)

(b² + c² – a²) tan A = 2bc cos A tan A 

= 2bc sin A

= 2kabc . . . . (1)

Also, cos B = (a² + c² – b²)/2ac

2ac cos B = (a² + c² – b²)  

(a² + c² – b²) tan B = 2ac cos B tan B

= 2ac sin B

= 2kabc . . . . (2)

Also, cos C = (a² + b² – c²)/2ab

2ab cos C = (a² + b² – c²)  . . . . (3)

(a² + b² – c²) tan C = 2ab cos C tan C

= 2ab sin C

= 2kabc . . . . (3)

From (1), (2) and (3), we get,

(c² + b² – a²) tan A = (a² + c² – b²) tan B = (a² + b² – c²) tan C

Hence, proved.

问题9.对于任何ΔABC，表明： $\frac{c-b\hspace{0.02cm}cosA}{b-c\hspace{0.02cm}cosA}=\frac{cosB}{cosC}$

解决方案：

According to sine rule in ΔABC,

a/sin A = b/sin B = c/sin C = k (constant) 

Here, L.H.S. = $\frac{c-b\hspace{0.02cm}cosA}{b-c\hspace{0.02cm}cosA}$

= $\frac{ksinC-ksinBcosA}{ksinB-ksinCcosA}$

= $\frac{ksinC-ksinBcosA}{ksinB-ksinCcosA}$

= $\frac{sin(π-(A+B))-sinBcosA}{sin(π-(A+C))-sinCcosA}$

= $\frac{sin(A+B)-sinBcosA}{sin(A+C)-sinCcosA}$

= $\frac{sinAcosB+cosAsinB-sinBcosA}{sinAcosC+cosAsinC-sinCcosA}$

= $\frac{sinAcosB}{sinAcosC}$

= $\frac{cosB}{cosC}$

= R.H.S.

Hence, proved.

问题10.对于任何ΔABC，表明：

a(cos B + cos C – 1)+ b(cos C + cos A – 1)+ c(cos A + cos B – 1)= 0

解决方案：

According to projection formula, we get,

a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A

Here, L.H.S. = a(cos B + cos C – 1) + b(cos C + cos A – 1) + c(cos A + cos B – 1)

= a cos B + a cos C – a + b cos C + b cos A – b + c cos A + c cos B – c

= c – b cos A + a cos C – a + a – c cos B + b cos A – b + b – a cos C + c cos B – c

= 0

= R.H.S.

Hence, proved.