Page 66 - mathsvol1ch1to3ans

66

a sin(B − C) R sin A sin(B − C)
=
2
2
2
b − c 2 R 2 sin B − sin C
sin(B + C) sin(B − C)
=
R sin(B + C) sin(B − C)
1
=
R
b sin(C − A)
a sin(B−C)
Because of symmetricity, we can easily show that = =
2
b −c 2 c − a 2
2
c sin(A − B) 1
= .
2
a − b 2 R
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a + b A + B A − B
(v) = tan cot .
a − b 2 2
Solution:
a + b R (sin A + sin B)
=
a − b R (sin A − sin B)
A + B A − B
2 sin cos
= 2 2
A − B A + B
2 sin cos
2 2
A + B A − B
   
sin cos
2 2
   
A + B A − B
=    
cos sin
2
2
A + B A − B
= tan cot
2 2
2
2
2
2
2
2
8. In a 4ABC, prove that (a − b + c ) tan B = (a + b − c ) tan C.
Solution:
tan B sin B cos C 2 sin B cos C
= =
tan C cos B sin C 2 cos B sin C
2
2
a + b − c 2
sin B
= 2ab
2
2
a + c − b 2
sin C
2ac
2
2
2
c sin B (a + b − c )
=
2
2
2
b sin C (a + c − b )
2
2
a + b − c 2
=
2
2
a + c − b 2
9. An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park
to be developed must be of maximum area. Find out the dimensions of the park.
Solution:
For a fixed perimeter, the equilateral triangle has the maximum area and the maximum area
s 2
is given by ∆ = √ sq.units. Here s = 120 m. Hence the area of the park is given by
3 3
√
14400 1600 3
∆ = √ = Clearly the sides of the park are equal and given by 40 m.
3 3 .