Question

of the three angles of a triangle the second angle is one third of the first angle and the third angle is 26 degree more than the first angle. find all the three angles of the triangle.
from the chapter linear equation in one one variable

Answer 1

Answer:

(Angle)1=66(Degree); (Angle)2=22(Degree); (Angle)3=92(Degree)

Step-by-step explanation:

A/C to Question,

(Angle)2=1/3(Angle)1

(Angle)3=(Angle)1+26(Degree)

Now,

By Angle Sum Property of Triangle,

(Angle)1+(Angle)2+(Angle)3=180(Degree)

(Angle)1+1/3(Angle)1+(Angle)1+26(Degree)=180(Degree)

2{(Angle)1}+1/3(Angle)1=154(Degree)

2+1/3{(Angle)1}=154(Degree)

7/3{(Angle)1}=154(Degree)

(Angle)1=(154x3)/7(Degree)

(Angle)1=66(Degree)

So,

(Angle)1=66(Degree)

(Angle)2=1/3(Angle)1=1/3x66(Degree)=22(Degree)

(Angle)3=(Angle)1+26(Degree)=66(Degree)+26(Degree)=92(Degree)

CONCLUSION:-

(Angle)1=66(Degree)

(Angle)2=22(Degree)

(Angle)3=92(Degree)

Answer 2

$\mathfrak{\underline{Answer:\:\angle A = 66,\:\angle B = 22, \:C =92}}$

$\rule{100}2$

$\textbf{\underline{\underline{Step-By-Step\:Explanation:}}}$

Let first angle be ' A ' , second be ' B ' and third be ' C ' .

It is stated that the second angle is one third of the first. So,  $\angle B =\dfrac{1}{3} \times {A}$

For angle C,

It is stated that the third angle is 26° more than the first angle, i.e $\angle C = A + 26$

→ Sum of all three interior angles of a triangle is 180°

Now,

$\sf \angle {A} + \angle {B} + \angle {C} = 180$

$\sf \implies {A} + \dfrac{1}{3} \times{A} + (A + 26) = {180}$

$\sf \implies \dfrac{7A}{3} + 26 = 180$

$\sf \implies \dfrac{7A}{3} = 180 - 26 \\\\ \implies \dfrac{7A}{3} = 154 \\\\ \implies {A} = 155\times \dfrac{3}{7} \\\\ \therefore {A} = 66$

Hence,

     ∠A = 66°

    $\angle B = \dfrac{1}{3} \times 66$ = 22°

    $\angle C = 66 + 26$ = 92°

$\rule{200}2$