Question

Find the sides of triangle if they are in ratio 5:2:3

Answer 1

Solution,

• Let the angles of a triangle be 2x,3x,5x.
• Sum of interior angles of a triangle is 180°
• Hence ∠A+∠B+∠C=180°
• ⇒2x+3x+5x=180°
• ⇒10x=180°
• ⇒x=18°
• Angles are 36°,54°,90

Required answer,

• Angles are 36°,54°,90

hope it's helpful to you

Answer 2

Given :-

• The sides of a triangle are in the ratio 5 : 2 :3.

To Find :-

• The sides of a triangle.

Concept used :-

Cosine law

• Let us consider a triangle ABC such that side AB = c units, side AC = b units and side BC = a units.

then

• angle betweem AB & AC is given by

$\rm :\longmapsto\:cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}$

Lets do the problem now !!

$\large\underline{\bold{Solution :- }}$

Let us consider a triangle ABC such that

• side AB = c units,

• side AC = b units

• side BC = a units.

It is given that

• The sides of triangle are in ratio 5:2:3

So,

• a : b : c = 5 : 2 : 3

So,

$\begin{gathered}\begin{gathered}\bf \: Let \: sides \: of \: triangle \: be - \begin{cases} &\sf{a \: = \: 5x} \\ &\sf{b \: = \: 2x}\\ &\sf{c \: = \: 3x} \end{cases}\end{gathered}\end{gathered}$

• Let A be the angle between the sides b and c,

then

$\rm :\longmapsto\:cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}$

$\rm :\longmapsto\:cosA = \dfrac{ {(2x)}^{2} + {(3x)}^{2} - {(5x)}^{2} }{2 \times 2x \times 3x}$

$\rm :\longmapsto\:cosA = \dfrac{ {4x}^{2} + {9x}^{2} - {25x}^{2} }{12 {x}^{2} }$

$\rm :\longmapsto\:cosA = - \: \dfrac{12 {x}^{2} }{12 {x}^{2} }$

$\rm :\longmapsto\:cosA \: = \: - 1$

$\rm :\implies\: \angle \: A \: = \: 180 \degree$

$\bf\implies \: \triangle \: is \: not \: possible$