Question

2. Three angles of a triangle (3x + 50), (x + 20) and (2x + 20). Greatest angle of the triangle is -​

Answer 1

In context to questions asked,

We have to determine the value of greatest angle of triangle.

As per questions,

It is given that,

Three angles of a triangle (3x + 50), (x + 20) and (2x + 20)

As we know that,

Sum of all angles of triangle is 180 degree.

So, we will get,

$(3x + 50) + (x + 20) + (2x + 20) = 180 \\ = > 6x + 90 = 180 \\ = > 6x = 180 - 90 \\ = > 6x = 90 \\ = > x = \frac{90}{6} \\ = > x = {15}^{0}$

Hence, the value of angles will be,

$(3x + 50) = 45 + 50 = {95}^{0} \\ (x + 20) = 15 + 20 = {35}^{0} \\ (2x + 20) = 30 + 20 = {50}^{0}$

Hence, value of greatest angle will be 95 degree

Answer 2

❒ Given :-

Three angles of a triangle (3x + 50), (x + 20) and (2x + 20)

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❒ To Find :-

Greatest angle of the triangle

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❒ Used Formula :-

$\bf \red{⟼ \: Sum \: of \: all \: angles _{(Triangle)}= 180°}$

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❒ Solution :-

In Δle,

Sum of all angles = 180°

Let,

First angle = (3x + 50)

Second angle = (x + 20)

Third angle = (2x + 20)

So,

$\bf ⟹ \: Sum \: of \: all \: angles _{(Triangle)}= 180°$

$\bf⟹ \: (3x + 50) + (x + 20) + (2x + 20) = 180 \degree$

$\bf⟹ \: 6x + 90 = 180 \degree$

$\bf⟹ \: 6x = 180 - 90 \degree$

$\bf⟹ \: 6x = 90 \degree$

$\displaystyle\bf⟹ \: x = \cancel \frac{90}{2}$

$\red{⟹ \: \underline{ \boxed{ \bf x = 15}}}$

Now,

First angle = (3x + 50) = 3×15 + 50 = 95°

Second angle = (x + 20) = 15 + 20 = 35°

Third angle = (2x + 20) = 2×15 + 20 = 50°

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❒ Verification :-

$\bf ⟹ \: Sum \: of \: all \: angles _{(Triangle)}= 180°$

$\bf ⟹ \:95 + 35 + 50= 180°$

$\bf ⟹ \: \boxed{ \bf180°= 180°}$

Hence, verified !

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