In a triangle ABC, Pand 1are …

We have,

According to the question,we are given that,
{tex}P Q \| B C{/tex}
Therefore, by thales theorem,
We have,
{tex}\frac { A P } { P B } = \frac { A Q } { Q C }{/tex}
{tex}\frac { 2.4 } { P B } = \frac { 2 } { 3 }{/tex}
{tex}\Rightarrow P B = \frac { 3 \times 2.4 } { 2 }{/tex}
{tex}= \frac { 3 \times 24 } { 20 }{/tex}
{tex}= \frac { 3 \times 6 } { 5 }{/tex}
{tex}= \frac { 18 } { 5 }{/tex}
{tex}\Rightarrow{/tex} PB = 3.6 cm
Now, AB = AP + PB
= 2.4 + 3.6
= 6 cm
In {tex}\triangle {/tex}APQ and  {tex}\triangle{/tex}ABC
{tex}\angle A = \angle A{/tex} [Common]
{tex}\angle A P Q = \angle A B C{/tex} [{tex}\because{/tex} PQ || BC {tex}\Rightarrow{/tex} Corresponding angles are equal]
{tex}\Rightarrow \triangle A P Q \sim \triangle A B C{/tex} [By AA criteria]
{tex}\Rightarrow \frac { A B } { A P } = \frac { B C } { P Q }{/tex} [Corresponding sides of similar triangles are proportional]
{tex}\Rightarrow \frac { 6 } { 2.4 } = \frac { 6 } { P Q }{/tex}
{tex}\Rightarrow P Q = \frac { 6 \times 2.4 } { 6 }{/tex}
{tex}\Rightarrow{/tex} PQ = 2.4 cm
Hence, AB = 6 cm and PQ = 2.4 cm