Answer:
[tex]\bold{\angle B = 51^\circ}[/tex]
Step-by-step explanation:
Given a [tex]\triangle ABC[/tex] such that [tex]\angle A[/tex] is the largest.
[tex]\angle A[/tex] is equal twice of [tex]\angle B[/tex]
[tex]\angle C[/tex] is 10 more than one third of [tex]\angle B[/tex]
To find:
Measurement of [tex]\angle B[/tex].
Solution:
Let [tex]\angle B =x^\circ[/tex]
As per question statement:
[tex]\angle A = 2\times \angle B=2x^\circ[/tex]
[tex]\angle C = \dfrac{1}{3}\angle B +10= \dfrac{1}{3}x +10 ^\circ[/tex]
Using the angle sum property of a triangle i.e. sum of all the three internal angles is equal to [tex]180^\circ[/tex].
[tex]2x+x+\dfrac{1}{3}x+10 = 180\\\Rightarrow \dfrac{10}{3}x =170\\\Rightarrow x = 51^\circ[/tex]
Therefore, the answer is:
[tex]\bold{\angle B = 51^\circ}[/tex]
