Question

of the three angles of a triangle one is twice the smallest and another one is thrice the smallest find the angles ​

Answer 1

Answer

$\boxed{\boxed{\mathsf{Measure\: of \:Angles \: are = {30}^{\circ} , {60}^{\circ} and {90}^{\circ}}}}$

Step-by-step Explanation :

Given :

• One of the three angles of a triangle is twice the smallest angle of the triangle .

• and the other angle is thrice the smallest angle .

To find :

• All the three angles .

Solution :

Now let,

$\longrightarrow \mathtt{The \: smallest \: angle \: be = X}$

Therefore According to Question :

It's said that one angles is twice the smallest angle and another angle is thrice the smallest angle . now since smallest angle = X we have,

$\longrightarrow \mathsf{Second \: angle \: = 2x}$

$\longrightarrow \mathsf{Third \: angle \: = 3x}$

Now we know that

$\boxed{\mathtt{\bigstar \; In\: \Delta ABC \: \longrightarrow \angle A + \angle B + \angle C = {180}^{\circ}}}$

Where

A , B and C are the internal angles of the triangle this property of triangle is known as Angle Sum Property Of Triangle

Therefore here we have ,

$\longrightarrow \: \mathsf{x + 2x + 3x = {180}^{\circ}}$

$\longrightarrow \: \mathsf{6x = {180}^{\circ}}$

*taking 6 to RHS we have*

$\longrightarrow \: \mathsf{x = \dfrac{{180}^{\circ}}{6}}$

$\longrightarrow \: \mathsf{x = \dfrac{{180}^{\circ}}{6}}$

$\longrightarrow \: \mathsf{x = {30}^{\circ}}$

Therefore now putting value of x in the assumed form of angle we have

$\longrightarrow \mathsf{Smallest \: angle \: = x = {30}^{\circ}}$

$\longrightarrow \mathsf{Second \: angle \: = 2x = 2(30) = {60}^{\circ}}$

$\longrightarrow \mathsf{Third \: angle \: = 3x = 3(30) = {90}^{\circ}}$

Therefore Required answer equal to

$\underline{\boxed{\mathsf{Measure\: of \:Angles \: are = {30}^{\circ} , {60}^{\circ} and {90}^{\circ}}}}$

Answer 2

Given :------

• second angle of ∆ is twice the smallest.
• third angle of ∆ is thrice the smallest .

Concept used :----

• Sum of all three angles of a ∆ is 180°.

$\pink{\bold{\underline{\underline{Solution(1)}}}}$ :-

Let the smallest angle be = x°

than, second angle = 2x°

third angle be = 3x°

A/q,

x + 2x + 3x = 180°

→ 6x = 180°

→ x = 180/6 = 30° = smallest angle

second angle = 2x = 2×30 = 60°

third angle = 3×30 = 90°

so, all three angles of ∆ are 30°,60° & 90° .

___________________________

$\red{\bold{\underline{\underline{solution(2)}}}}$

Let the required ratio of Smallest angle : second angle : third angle be = 1 : 2 : 3

Total = 1 + 2 + 3 = 6 units.

and sum of angles of ∆ we know = 180°

That means ,

6 unit ------------- 180°

1 unit -------------- 30° (smallest angle)

2 unit ------------- 60° (second angle)

3 unit ------------- 90° (Third angle)

_______________________________________

$\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

→ sum of any two sides of ∆ is larger than third sides .

→ if a ∆ is right angled ∆ , we can use pythagoras theoram ,

(Hpyotenuse)² = (perpendicular)² + (Base)²

→ Area of Right angled ∆ = 1/2 × Base × Height

→ Sides ratio of Right angled ∆ is = 1 : √3 : 2 ( opposite of 30°: 60: 90° )

$\mathcal{BE\:\:BRAINLY}$