**Question 1. In a ∆ABC, if a = 5, b = 6 and C = 60**^{o}, show that its area is (15√3)/2 sq. units.

^{o}, show that its area is (15√3)/2 sq. units.

**Solution:**

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.

**Question 2. In a ∆ABC, if a = √2, b = √3 and c = √5 show that its area is √6/2 sq. units.**

**Solution:**

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

According to Cosine formula, cos C = (a

^{2}+ b^{2 }– c^{2})/2ab= (2 + 3 – 5)/2√6

= 0

So, sin C = √(1–cos

^{2}C) = 1Now 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.

**Question 3. The sides of a triangle are a = 4, b = 6 and c = 8, show that: **

**8 cos A + 16 cos B + 4 cos C = 17**

**Solution:**

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

According to Cosine formula, cos A = (b

^{2}+ c^{2}– a^{2})/2bc= (36 + 64 – 16)/96

= 84/96

= 7/8

Also, cos B = (a

^{2}+ c^{2}– b^{2})/2ac= (16 + 64 – 36)/64

= 44/64

= 11/16

Also, cos C = (a

^{2}+ b^{2}– c^{2})/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.

**Question 4. In a ∆ABC, if a = 18, b = 24, c = 30, find cos A, cos B and cos C.**

**Solution: **

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

According to Cosine formula, cos A = (b

^{2}+ c^{2}– a^{2})/2bc= (24

^{2}+ 30^{2}– 18^{2})/2(24)(30)= (576 + 900 – 324)/1440

= 1152/1440

= 4/5

Also, cos B = (a

^{2}+ c^{2}– b^{2})/2ac= (324 + 900 – 576)/2(18)(30)

= 648/1080

= 3/5

Also, cos C = (a

^{2 }+ b^{2}– c^{2})/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.

**Question 5. For any **Δ**ABC, show that b (c cos A – a cos C) = c**^{2} – a^{2}

^{2}– a

^{2}

**Solution:**

According to Cosine formula,

cos A = (b

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

^{2}+ c^{2}– a^{2})/2 . . . . (1)Also, cos C = (a

^{2}+ b^{2}– c^{2})/2ab=> ab cos C = (a

^{2}+ b^{2}– c^{2})/2 . . . . (2)Subtracting (2) from (1), we get,

L.H.S. = bc cos A – ab cos C = (b

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

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

^{2}– 2a^{2})/2= c

^{2}– a^{2}= R.H.S.

Hence, proved.

**Question 6. For any **Δ**ABC show that c (a cos B – b cos A) = a**^{2} – b^{2}

^{2}– b

^{2}

**Solution:**

According to Cosine formula,

cos B = (a

^{2}+ c^{2}– b^{2})/2ac=> ac cos B = (a

^{2}+ c^{2}– b^{2})/2 . . . . (1)Also cos A = (b

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

^{2}+ c^{2}– a^{2})/2 . . . . (2)Subtracting (2) from (1), we get,

L.H.S. = ac cos B – bc cos A = (a

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

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

^{2}– 2b^{2})/2= a

^{2}– b^{2}= R.H.S.

Hence, proved.

**Question 7. For any **Δ**ABC show that 2 (bc cos A + ca cos B + ab cos C) = a**^{2} + b^{2} + c^{2}

^{2}+ b

^{2}+ c

^{2}

**Solution:**

According to Cosine formula,

cos A = (b

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

^{2}+ c^{2}– a^{2}. . . . (1)Also, cos B = (a

^{2}+ c^{2}– b^{2})/2ac=> 2ac cos B = a

^{2}+ c^{2}– b^{2}. . . . (2)Also, cos C = (a

^{2}+ b^{2}– c^{2})/2ab=> 2ab cos C = a

^{2}+ b^{2}– c^{2}. . . . (3)Adding (1), (2) and (3), we get,

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

= b

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

^{2}+ b^{2}+ c^{2}= R.H.S.

Hence, proved.

**Question 8. For any **Δ**ABC show that (c**^{2 }+ b^{2 }– a^{2}) tan A = (a^{2} + c^{2 }– b^{2}) tan B = (a^{2 }+ b^{2 }– c^{2}) tan C

^{2 }+ b

^{2 }– a

^{2}) tan A = (a

^{2}+ c

^{2 }– b

^{2}) tan B = (a

^{2 }+ b

^{2 }– c

^{2}) tan C

**Solution:**

According to sine rule in ΔABC,

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

According to Cosine formula,

cos A = (b

^{2}+ c^{2}– a^{2})/2bc2bc cos A = (b

^{2}+ c^{2}– a^{2})(b

^{2}+ c^{2}– a^{2}) tan A = 2bc cos A tan A= 2bc sin A

= 2kabc . . . . (1)

Also, cos B = (a

^{2}+ c^{2}– b^{2})/2ac2ac cos B = (a

^{2}+ c^{2}– b^{2})(a

^{2}+ c^{2}– b^{2}) tan B = 2ac cos B tan B= 2ac sin B

= 2kabc . . . . (2)

Also, cos C = (a

^{2}+ b^{2}– c^{2})/2ab2ab cos C = (a

^{2}+ b^{2}– c^{2}) . . . . (3)(a

^{2}+ b^{2}– c^{2}) tan C = 2ab cos C tan C= 2ab sin C

= 2kabc . . . . (3)

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

(c

^{2}+ b^{2}– a^{2}) tan A = (a^{2}+ c^{2}– b^{2}) tan B = (a^{2}+ b^{2}– c^{2}) tan C

Hence, proved.

**Question 9. For any ΔABC show that:**

**Solution:**

According to sine rule in ΔABC,

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

Here, L.H.S. =

=

=

=

=

=

=

=

= R.H.S.

Hence, proved.

**Question 10. For any **Δ**ABC show that: **

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

**Solution:**

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.