**Question 11. In any ∆ABC, prove the following: a cos A + b cos B + c cos C = 2b sin A sin C.**

**Solution:**

According to sine rule in ΔABC,

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

L.H.S. = a cos A + b cos B + c cos C

= k sin A cos A + k sin B cos B + k sin C cos C

= (k/2) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]

= (k/2) [sin 2A + sin 2B + sin 2C]

= (k/2) [sin 2A + sin 2B + sin 2C]

= (k/2) [sin(A+B) cos(A–B) + sin C cos C]

= (k/2) [sin(π–C) cos(A–B) + sin C cos(π–(A+B))]

= (k/2) [sin C cos(A–B) + sin C cos(A+B)]

= k sin C [2 sin A sin B]

= 2 (sin A) (k sin B) (sin C)

= 2b sin A sin C

= R.H.S.

Hence, proved.

**Question 12. Prove that a**^{2} = (b+c)^{2} – 4bc cos^{2}A/2

^{2}= (b+c)

^{2}– 4bc cos

^{2}A/2

**Solution:**

According to Cosine rule,

=> cos A = (b

^{2}+ c^{2}– a^{2})/2bc=> 2bc cos A = b

^{2}+ c^{2}– a^{2}=> a

^{2}= b^{2}+ c^{2}– 2bc cos A=> a

^{2}= b^{2}+ c^{2}– 2bc (2cos^{2}A/2 – 1)=> a

^{2}= b^{2}+ c^{2}– 4bc cos^{2}A/2 + 2bc=> a

^{2}= (b+c)^{2}– 4bc cos^{2}A/2

Hence, proved.

**Question 13. Prove that 4(bc cos**^{2}A/2 + ac cos^{2}B/2 + ab cos^{2}C/2) = (a + b + c)^{2}

^{2}A/2 + ac cos

^{2}B/2 + ab cos

^{2}C/2) = (a + b + c)

^{2}

**Solution:**

According to Cosine rule,

cos A = (b

^{2}+ c^{2}– a^{2})/2bc . . . . (1)cos B = (a

^{2}+ c^{2}– b^{2})/2ac . . . . (2)cos C = (a

^{2}+ b^{2}– c^{2})/2ab . . . . (3)We have,

L.H.S. = 4(bc cos

^{2}A/2 + ac cos^{2}B/2 + ab cos^{2}C/2)= 2(2bc cos

^{2}A/2 + 2ac cos^{2}B/2 + 2ab cos^{2}C/2)= 2[bc(1–cos A) + ac(1–cos B) + ab(1–cos C)]

= 2bc – 2bc cos A + 2ac – 2ac cos B + 2ab – 2ab cos C

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

L.H.S. = 2bc – [b

^{2}+ c^{2}– a^{2}] + 2ac – [a^{2}+ c^{2}– b^{2}] + 2ab – [a^{2}+ b^{2}– c^{2}]= 2bc – b

^{2}– c^{2}+ a^{2}+ 2ac – a^{2}– c^{2}+ b^{2}+ 2ab – a^{2}– b^{2}+ c^{2}= a

^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca= (a + b + c)

^{2}= R.H.S.

Hence, proved.

**Question 14. In a** Δ**ABC, prove that sin**^{3}A cos(B–C) + sin^{3}B cos(C–A)+ sin^{3}C cos(A–B) = 3 sinA sinB sin C

^{3}A cos(B–C) + sin

^{3}B cos(C–A)+ sin

^{3}C cos(A–B) = 3 sinA sinB sin C

**Solution:**

We have,

L.H.S. = sin

^{3}A cos(B–C) + sin^{3}B cos(C–A)+ sin^{3}C cos(A–B)= sin

^{2}A sin A cos(B–C) + sin^{2}B sin B cos(C–A)+ sin^{2}C sin C cos(A–B)= sin

^{2}A sin(π–(B+C)) cos(B–C) + sin^{2}B sin(π–(A+C)) cos(C–A)+ sin^{2}C sin(π–(A+B)) cos(A–B)= sin

^{2}A sin(B+C) cos(B–C) + sin^{2}B sin(A+C) cos(C–A)+ sin^{2}C sin(A+B) cos(A–B)= sin

^{2}A [sin2B + sin2C] + sin^{2}B [sin2A + sin2C]+ sin^{2}C [sin2A + sin2B]= sin

^{2}A [2sinB cosB + 2sinC cosC] + sin^{2}B [2sinA cosA+ 2sinC cosC]+ sin^{2}C [2sinA cosA + 2sinB cosB]= 2sin

^{2}A sinB cosB + 2sin^{2}A sinC cosC + 2sin^{2}B sinA cosA + 2sin^{2}B sinC cosC + 2sin^{2}C sinA cosA + 2sin^{2}C sinB cosBAccording to sine rule in ΔABC,

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

So, L.H.S. becomes,

= 2a

^{2}k^{2}(bk) (cosB) + 2a^{2}k^{2}(ck) (cosC) + 2b^{2}k^{2}(ak) (cosA) + 2b^{2}k^{2}(ck) (cosC) + 2c^{2}k^{2}(ak) (cosA) + 2c^{2}k^{2}(bk) cosB= abk

^{3}(a cosB + b cosA) + ack^{3}(a cosC + c cosA) + bck^{3}(c cosB + b cosC)= abck

^{3 }+ abck^{3}+ abck^{3}= 3abck

^{3}= 3 (ak) (bk) (ck)

= 3 sin A sin B sin C

= R.H.S.

Hence, proved.

**Question 15. In any **Δ**ABC, (b+c)/12 = (c+a)/13 = (a+b)/15, prove that (cosA)/2 = (cosB)/7 = (cosC)/11.**

**Solution:**

We are given,

(b+c)/12 = (c+a)/13 = (a+b)/15 = k (say)

=> b + c = 12k . . . . (1)

=> c + a = 13k . . . . (2)

=> a + b = 15k . . . . (3)

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

=> b + c + c + a + a + b = 12k + 13k + 15k

=> 2a + 2b + 2c = 40k

=> a + b + c = 20k . . . . (4)

From (1), (2), (3) and (4), we get,

=> a = 8k, b = 7k and c = 5k

According to Cosine formula,

cos A = (b

^{2}+ c^{2}– a^{2})/2bc= (49k

^{2}+ 25k^{2}– 64k^{2})/70k^{2}= 10/70

= 1/7

cos B = (a

^{2}+ c^{2}– b^{2})/2ac= (64k

^{2}+ 25k^{2}– 49k^{2})/80k^{2}= 40/80

= 1/2

cos C = (a

^{2}+ b^{2}– c2)/2ab= (64k

^{2}+ 49k^{2}– 25k^{2})/112k^{2}= 88/112

= 11/14

Hence, cosA : cosB : cosC = (1/7) : (1/2) : (11/14) = 2 : 7 : 11

=> (cosA)/2 = (cosB)/7 = (cosC)/11

Hence, proved.

**Question 16. In **Δ**ABC, if ∠B = 60**^{o}, prove that (a + b + c) (a – b + c) = 3ca.

^{o}, prove that (a + b + c) (a – b + c) = 3ca.

**Solution:**

We have, ∠B = 60

^{o}According to Cosine formula,

cosB = (a

^{2}+ c^{2}– b^{2})/2ac=> cos 60

^{o}= (a^{2}+ c^{2}– b^{2})/2ac=> 1/2 = (a

^{2}+ c^{2}– b^{2})/2ac=> a

^{2}+ c^{2}– b^{2}= ac . . . . (1)Now L.H.S. = (a + b + c) (a – b + c)

= a

^{2}– ab + ac + ab – b^{2}+ bc + ca – bc + c^{2}= a

^{2}+ c^{2}– b^{2}+ 2ac= ac + 2ac

= 3ac

= R.H.S.

Hence, proved.

**Question 17. If in a **Δ**ABC, cos**^{2}A + cos^{2}B + cos^{2}C = 1, prove that the triangle is right angled.

^{2}A + cos

^{2}B + cos

^{2}C = 1, prove that the triangle is right angled.

**Solution:**

We are given,

=> cos

^{2}A + cos^{2}B + cos^{2}C = 1=> cos

^{2}A + cos^{2}B = 1 – cos^{2}C=> cos

^{2}A + cos^{2}B = sin^{2}C=> cos

^{2}A = sin^{2}C − cos^{2}B=> −cos(B+C) cos(B−C) = cos

^{2}A=> −cos(π−A) cos(B−C) = cos

^{2}A=> cos

^{2}A − cosA cos(B−C) = 0=> cosA (cosA − cos(B−C)) = 0

=> cosA [cos(π−(B+C)) − cos(B−C)] = 0

=> cosA [−cos(B+C) − cos(B−C)] = 0

=> cosA [−2 cosB cosC] = 0

=> cosA cos B cos C = 0

=> cosA = 0 or cosB = 0 or cosC = 0

=> A = 90

^{o}or B = 90^{o}or C = 90^{o}

Therefore, the triangle is right-angled.

**Question 18. In a **Δ**ABC, if cosC = sinA/2sinB, prove that the triangle is isosceles. **

**Solution:**

Here, we are given

=> cosC = sinA/2sinB

=> 2 sinB cosC = sinA

=> 2 (kb) [(a

^{2}+b^{2}−c^{2})/2ab] = ka=> a

^{2}+ b^{2}− c^{2}= a^{2}=> b

^{2}= c^{2}=> b = c

Therefore, the triangle is isosceles.

**Question 19. Two ships leave a port at the same time. One goes 24 km/hr in the direction N38**^{o}E and other travels 32 km/hr in the direction S52^{o}E. Find the distance between the ships at the end of 3 hours.

^{o}E and other travels 32 km/hr in the direction S52

^{o}E. Find the distance between the ships at the end of 3 hours.

**Solution:**

Let A be the point from where ships leave. AB is the distance travelled by one and AC is the distance travelled by the second in 3 hours.

We have to find BC, the distance between the ships at the end of 3 hours. Here,

AB = 3(24) = 72 km and AC = 3(32) = 96 km.

Using Cosine formula in ΔABC, we get,

BC

^{2}= AB^{2}+ AC^{2}+ 2 (AB) (AC) cos90^{0}BC

^{2}= 72^{2}+ 96^{2}BC

^{2}= 14400BC = 120 km

Therefore, the distance between the ships at the end of 3 hours is 120 km.