Question
In a $\triangle\text{ABC},\angle\text{A}=\text{x}^\circ,\angle\text{B}=3\text{x}^\circ$ and $\angle\text{C}=\text{y}^\circ$ if 3y - 5x = 30, , prove that the triangle is right angled.
✓
Answer
We have,
$\angle\text{A}=\text{x}^\circ\ ....(\text{i})$
$\angle\text{B}=3\text{x}^\circ\ .....(\text{ii})$
And, $\angle\text{C}=\text{y}^\circ\ ......(\text{iii})$
We know that, the sum of angles of a triangle is 180°
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
x + 3x + y = 180 [using (i), (ii) and (iii)]
⇒ 4x + y = 180 .....(iv)
Now, 3y - 5x = 30 .....(v) [given]
Multiplying equation (iv) by 3 we get
12x + 3y = 540 .....(vi)
Subtracting equation (v) from equation (vi) we get
12x + 5x = 540 - 30
⇒ 17x = 510
$\Rightarrow\text{x}=\frac{510}{17}$
⇒ x = 30
Putting x = 30 in equation (iv) we get
4 × 30 + y = 180
⇒ 120 + y = 180
⇒ y = 180 - 120
⇒ y = 60
Now, $\angle\text{B}=3\text{x}^\circ$
$\Rightarrow\angle\text{B}=3\times30^\circ$
$\Rightarrow\angle\text{B}=90^\circ$
$\therefore\triangle\text{ABC}$ is the right angle triangle.
Hence prived.
$\angle\text{A}=\text{x}^\circ\ ....(\text{i})$
$\angle\text{B}=3\text{x}^\circ\ .....(\text{ii})$
And, $\angle\text{C}=\text{y}^\circ\ ......(\text{iii})$
We know that, the sum of angles of a triangle is 180°
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
x + 3x + y = 180 [using (i), (ii) and (iii)]
⇒ 4x + y = 180 .....(iv)
Now, 3y - 5x = 30 .....(v) [given]
Multiplying equation (iv) by 3 we get
12x + 3y = 540 .....(vi)
Subtracting equation (v) from equation (vi) we get
12x + 5x = 540 - 30
⇒ 17x = 510
$\Rightarrow\text{x}=\frac{510}{17}$
⇒ x = 30
Putting x = 30 in equation (iv) we get
4 × 30 + y = 180
⇒ 120 + y = 180
⇒ y = 180 - 120
⇒ y = 60
Now, $\angle\text{B}=3\text{x}^\circ$
$\Rightarrow\angle\text{B}=3\times30^\circ$
$\Rightarrow\angle\text{B}=90^\circ$
$\therefore\triangle\text{ABC}$ is the right angle triangle.
Hence prived.
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