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10. A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find
the dimensions of the triangle so formed.

Solution:

Considering the length of rope as perimeter,the largest area of the triangle formed by this rope is
s 2
an equilateral triangle. The largest area formed is given by ∆ = √ sq.units. Here s = 12 m.
3 3
144 √
Hence the area is given by ∆ = √ = 16 3. Clearly the sides are equal and given by 4 m.
3 3
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11. Derive Projection formula from (i) Law of sines, (ii) Law of cosines.

Solution:
Projection Formula gives the relation between angles and sides of a triangle. We can find the length
of a side of the triangle if other two sides and corresponding angles are given using projection
formula. If a, b and c be the length of sides of a triangle and A, B and C are angles opposite to the
sides respectively, then projection formula is given below:
a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A

Let us prove the first one using law of cosines.
2
2
2
a = b + c + 2bc cos(A)
2
2
= b + c + 2bc cos(B + C)
2
2
= b + c + 2bc (cos B cos C − sin B sin C) − (b sin C − c sin B) 2
2
2
2
2
2
2
= b + c + 2bc cos B cos C − 2bc sin B sin C − b sin C − c sin B + 2bc sin B sin C
2
2
2
2
2
2
= b + c + 2bc cos B cos C − b sin C − c sin B

2
2

= b 2 1 − sin C + c 2 1 − sin B + 2bc cos C cos B
2
2
2
2
= b cos C + c cos B + 2bc cos C cos B
= (b cos C + c cos B) 2
Taking square root on both sides, the result follows. Let us prove using Law of Sines
a cos B + b cos A = R sin A cos B + R sin B cos A
= R sin(A + B)
= R sin C
= c
Exercise 3.10: