Question

Topic: Triangles

Solve!​

Answer 1

$\large\underline{\sf{Solution-1}}$

We know,

In triangle ABC, sides of a triangle is represented as

$\rm :\longmapsto\:AB = c$

$\rm :\longmapsto\:BC = a$

$\rm :\longmapsto\:CA = b$

Now, Given that

$\boxed{ \bf \: { In \: \: \triangle \: \: ABC}}$

$\red{\rm :\longmapsto\:A = 2B}$

and

$\red{\rm :\longmapsto\:b = c}$

$\rm :\implies\:AC = AB$

$\rm :\implies\: \angle \:B \: = \: \angle \:C$

[ Angle opposite to equal sides are always equal ]

Now, we know that,

Sum of all interior angles of a triangle is supplementary.

$\rm :\longmapsto\:\angle \:A + \angle \:B + \angle \:C = 180 \degree$

$\rm :\longmapsto\:2\angle \:B + \angle \:B + \angle \:B = 180 \degree$

$\rm :\longmapsto\:4\angle \:B = 180 \degree$

$\rm :\longmapsto\:\angle \:B = 45 \degree$

So,

$\rm :\longmapsto\:\angle \:C = \angle \:B = 45 \degree$

and

$\rm :\longmapsto\:\angle \:A =2 \angle \:B = 90 \degree$

So,

$\bf\implies \triangle ABC \: is \: right \: angled \: \triangle \:$

So, Statement is incorrect.

$\large\underline{\sf{Solution-2}}$

Given that,

$\red{\rm :\longmapsto\:In \: \: \triangle \: \: ABC}$

$\rm :\longmapsto\:\angle \:A = 40\degree \:$

$\rm :\longmapsto\:\angle \:B = 65\degree \:$

We know,

Sum of all interior angles of a triangle is supplementary.

So,

$\rm :\longmapsto\:\angle \:A + \angle \:B + \angle \:C = 180 \degree$

$\rm :\longmapsto\:40\degree + 65\degree + \angle \:C = 180\degree$

$\rm :\longmapsto\:105\degree + \angle \:C = 180\degree$

$\rm :\longmapsto\:\angle \:C = 180\degree - 105\degree$

$\rm :\longmapsto\:\angle \:C = 75\degree$

Now, We know that, By Sine Law

$\red{\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}}$

$\red{\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{c}{sinC}}$

$\rm :\longmapsto\:\dfrac{a}{c} = \dfrac{sinA}{sinC}$

$\rm :\longmapsto\:\dfrac{a}{c} = \dfrac{sin40\degree}{sin75\degree}$

$\rm :\longmapsto\:\dfrac{a}{c} = sin40\degree \: cosec75\degree$

• Hence, Statement is correct.

Additional Information :-

$\boxed{ \rm{ cosA = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}}}$

$\boxed{ \rm{ cosB = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ca}}}$

$\boxed{ \rm{ cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}}}$

$\boxed{ \rm{ a = bcosC + CcosB}}$

$\boxed{ \rm{ b = acosC + ccosA}}$

$\boxed{ \rm{ c = acosB + bcosA}}$