If the lengths of the sides of a triangle are in A.P.

Question

If the lengths of the sides of a triangle are in A.P.
and the greatest angle is double the smallest, then
a ratio of lengths of the sides of this triangle is :

Accepted Answer

Let smallest angle $$\angle A$$ = $$\theta $$

and largest angle $$\angle C$$ is double of $$\angle A$$.

$$ \therefore $$ $$\angle C$$ = 2$$\theta $$

$$ \therefore $$ $$\angle B$$ = $$\pi $$ - 3$$\theta $$

Let length of sides are

a, b, c

where a < b < c

As a, b, c are in A.P then

2b = a + c

From sin rule we can say,

2 sin B = sin A + sin C

$$ \Rightarrow $$ 2 sin ($$\pi $$ -- 3$$\theta $$) = sin $$\theta $$ + sin 2$$\theta $$

$$ \Rightarrow $$ 2 sin (3$$\theta $$) = sin $$\theta $$ + sin 2$$\theta $$

$$ \Rightarrow $$ 2(3sin $$\theta $$ - 4sin3 $$\theta $$) = sin$$\theta $$ + 2sin $$\theta $$ cos $$\theta $$

$$ \Rightarrow $$ 2(3 - 4sin2 $$\theta $$) = 1 + 2cos $$\theta $$

$$ \Rightarrow $$ 2(3 - 4 + 4cos2 $$\theta $$) = 1 + 2cos $$\theta $$

$$ \Rightarrow $$ 2(4cos2 $$\theta $$ - 1) = 1 + 2cos $$\theta $$

$$ \Rightarrow $$ 8 cos2 $$\theta $$ -- 2 cos $$\theta $$ -- 3 = 0

$$ \Rightarrow $$ 8 cos2 $$\theta $$ -- 6 cos $$\theta $$ + 4 cos $$\theta $$ -- 3 = 0

$$ \Rightarrow $$ (2 cos $$\theta $$ + 1) ( 4 cos $$\theta $$ - 3) = 0

$$ \Rightarrow $$ cos $$\theta $$ = $${3 \over 4}$$ or cos $$\theta $$ = $$-{1 \over 2}$$

But cos $$\theta $$ = $$-{1 \over 2}$$ is not possible as $$\angle C$$ is acute angle.

Now we have to find,

a : b : c

$$ \Rightarrow $$ sin A : sin B : sin C [ from sin rule]

$$ \Rightarrow $$ sin $$\theta $$ : sin ($$\pi $$ - 3$$\theta $$)  : sin 2$$\theta $$

$$ \Rightarrow $$ sin $$\theta $$ : sin 3$$\theta $$  : sin 2$$\theta $$

$$ \Rightarrow $$ sin $$\theta $$ : (3sin $$\theta $$ - 4sin3 $$\theta $$)  : 2sin $$\theta $$ cos $$\theta $$

$$ \Rightarrow $$ 1 : (3 - 4sin2 $$\theta $$)  : 2 cos $$\theta $$

$$ \Rightarrow $$ 1 : (3 - 4 + 4cos2 $$\theta $$)  : 2 cos $$\theta $$

$$ \Rightarrow $$ 1 : (4cos2 $$\theta $$ - 1)  : 2 cos $$\theta $$

$$ \Rightarrow $$ 1 : (4$${\left( {{3 \over 4}} \right)^2}$$ - 1)  : 2 $$ \times $$ $${{3 \over 4}}$$

$$ \Rightarrow $$ 1 : $${5 \over 4}$$  : $${{3 \over 2}}$$

$$ \Rightarrow $$ 4 : 5  : 6

Solutions · 1