Question

Find x, if the angles of a triangle are :
(ii) x°, 2x°, 2x
(iii) 2x, 4X, 6x
(tell how you did it with workings)​

Answer 1

Answer :-

• x in (i) is 36°.
• x in (ii) is 15°

Given :-

Angles of the triangle :-

1. (ii) x°, 2x°, 2x
2. (iii) 2x, 4X, 6x

To find :-

• The value of x.

Solution :-

We know that,

• Sum of angles of the triangle = 180°.

(i) x + 2x + 2x = 180

➝ 5x = 180

➝ x = 180/5

➝ x = 36°.

Therefore, Angles of the triangle are :-

1. 1st angle = 36°.
2. 2nd angle = 72°.
3. 3rd angle = 72°.

(ii) 2x + 4x + 6x = 180°

➝ 12x = 180°

➝ x = 180/12$.$

➝ x = 15°

Therefore, Angles of the triangle are :-

1. 1st angle = 30°.(x × 2)
2. 2nd angle = 60°.(x × 4)
3. 3rd angle = 90°.(x × 6)

Answer 2

AnswEr-:

1. $\longrightarrow {\mathrm{The\:value \:of\:x\; = 36^{⁰}}}$
2. $\longrightarrow {\mathrm{The\:value \:of\:x\; = 15^{⁰}}}$

Explanation-:

$\mathrm{Given-:}$

• Interior angles of triangles-:

• (ii) x°, 2x°, 2x⁰
• (iii) 2x⁰, 4x⁰, 6x⁰

$\mathrm{To\: Find-:}$

• The Value of x .

$\dag {\mathrm {Solution \:of\:Question \:-:}}$

As, We know that,

• $\dag{\underline {\mathrm{Total \:Sum\:of\:interior \:angle \:of\:Triangle\:is\:180^{⁰}}}}$

Or ,

• $\dag{\underline {\mathrm{\angle A + \angle B + \angle C = 180^{⁰}}}}$ ______[ Formula 1]

________________________________

• Solution of (i) -:

• Angle A = x⁰

• Angle B = 2x⁰

• Angle C = 2x⁰

• Now , By putting known Values in Formula 1 or Formula of sum of angle of triangle-:

• $\longrightarrow {\mathrm{x^{⁰} + 2x^{⁰} + 2x^{⁰} = 180^{⁰}}}$

• $\longrightarrow {\mathrm{ 5x^{⁰} = 180^{⁰}}}$

• $\longrightarrow {\mathrm{ x = \dfrac{\cancel {180}}{\cancel {5}}}}$

• $\longrightarrow {\mathrm{ x = 36^{⁰}}}$

Therefore ,

• $\longrightarrow {\mathrm{The\:value \:of\:x\; = 36^{⁰}}}$

Putting x = 36⁰ -:

• Angle A = x⁰ = 36 ⁰
• Angle B = 2x⁰ = 2 × 36⁰ = 72⁰

• Angle C = 2x⁰ = 2 × 36⁰ = 72⁰

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• Solution of (ii) -:

• Angle A = 2x⁰

• Angle B = 4x⁰

• Angle C = 6x⁰

Now , By putting known Values in Formula 1 or Formula of sum of angle of triangle-:

• $\longrightarrow {\mathrm{2x^{⁰} + 4x^{⁰} + 6x^{⁰} = 180^{⁰}}}$

• $\longrightarrow {\mathrm{ 12x^{⁰} = 180^{⁰}}}$

• $\longrightarrow {\mathrm{ x = \dfrac{\cancel {180}}{\cancel {12}}}}$

• $\longrightarrow {\mathrm{ x = 15^{⁰}}}$

Therefore ,

• $\longrightarrow {\mathrm{The\:value \:of\:x\; = 15^{⁰}}}$

Putting x = 15⁰ -:

• Angle A = 2x⁰ = 2 × 15 = 30⁰

• Angle B = 4x⁰ = 4 × 15⁰ = 60⁰

• Angle C = 6x⁰ = 6 × 36⁰ = 90⁰

_______________________________

Hence ,

1. $\longrightarrow {\mathrm{The\:value \:of\:x\; = 36^{⁰}}}$
2. $\longrightarrow {\mathrm{The\:value \:of\:x\; = 15^{⁰}}}$

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