Test Index

Properties of Triangles

Section: Mathematics

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Question : 2 of 50

Marks: +1, -0

In a triangle

A B C

a, b, c

are in arithmetic progression and the angle

A

is twice the angle C , then

\cos A:\cos B:\cos C=

[AP EAPCET 24 May 2025 S1]

$2:3:4$
$3:4:8$
$2:9:12$
$1:9:6$

Solution:

Since,

a, b

and

c

are in AP

So,

2 b=a+c

Also,

\angle A=2\angle C\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)

Now, in a

\triangle A B C

,

a=2 R \sin A, b=2 R \sin B, c=2 R \sin C

So,

2 b=a+c

\;\Rightarrow\;\; 2 \cdot 2 R \sin B=2 R \sin A+2 R \sin C

\;\Rightarrow\;\; 2 \sin B= \sin A+ \sin C

In a

\triangle A B C

,

\;A+B+C=180^{\circ}

\Rightarrow 2 C+B+C=180^{\circ}

\Rightarrow B=180^{\circ}-3 C

So,

\sin B = \sin (180^{\circ}-3 C)

= \sin (3C) \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)

And

2\sin B = \sin A + \sin C

\Rightarrow 2\sin (3C) = \sin (2C) + \sin C

[using Eq. (i) and (ii)]

=2\sin C \cos C + \sin C

\;\;\text{ and }\; \sin (3C)=3\sin C-4\sin^3 C

\;\;\text{ So, }\; 2\sin (3C)=2(3-4\sin^2 C)

=2\cos C+1

=2(4\cos^2 C-1)

\Rightarrow 8\cos^2 C-2=2\cos C+1

\Rightarrow 8\cos^2 C-2\cos C-3=0

\therefore \cos C = \frac{2 \pm \sqrt{4-4(8)(-3)}}{16}

= \frac{2 \pm \sqrt{100}}{16} = \frac{2 \pm 10}{16}

So,

\cos C = \frac{3}{4}, \cos C = -\frac{1}{2}

Since,

C

must be positive,

So,

\cos C = \frac{3}{4}

Now,

\cos A = \cos (2C)

= 2\cos^2 C-1 = 2\left(\frac{3}{4}\right)^2-1 = \frac{1}{8}

And

B=180^{\circ}-3C

\cos B = \cos (180^{\circ}-3C)

= -\cos (3C)

= -(4\cos^3 C - 3\cos C)

= -\left(4\left(\frac{27}{64}\right) - 3\cdot \frac{3}{4}\right)

= -\left(\frac{27}{16} - \frac{9}{4}\right)

= -\left(\frac{27-36}{16}\right) = \frac{9}{16}

\therefore \cos B = \frac{9}{16}

So,

\cos A : \cos B : \cos C

= \frac{1}{8} : \frac{9}{16} : \frac{3}{4}

= \frac{2:9:12}{16}

So,

\cos A : \cos B : \cos C = 2:9:12

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