Question

In a $\Delta ABC$, $\angle C = 3\angle B = 2(\angle A + \angle B)$. Find the three angles.

Answer 1

Hint: Let us consider $\angle A = x,$ and $\angle B = y$
We can apply the concept that the sum of three angles of a triangle is $180_{}^o$.

Complete step-by-step answer:
It is given that$\angle C = 3\angle B = 2(\angle A + \angle B)$,
Therefore, we can write that $\angle C = 3y = 2(x + y)$ as we have supposed $\angle B = y,\angle A = x$
Thus,
$3y = 2(x + y)$
Multiply the value we get,
$3y = 2x + 2y$
Taking the \[y\]terms in LHS we get,
$3y - 2y = 2x$
On solving we get,
$y = 2x$
Since $\angle B = y$
∴ $\angle B = 2x$
$\angle C = 3\angle B = 3.2x = 6x$
As we all know that the sum of the angles of a triangle is $180_{}^o$
Therefore, we can write that
$\angle A + \angle B + \angle C = 180_{}^o$
Substitute the values we get,
$x + 2x + 6x = 180_{}^o$
On adding the terms, we get
$9x = 180_{}^o$
Dividing the terms we get,
$x = 20_{}^o$
So, as we have taken $\angle A = x$
Value of $\angle A = x = 20_{}^o$
Value of $\angle B = 2x = 2.20_{}^o = 40_{}^o$
Value of $\angle C = 6x = 6.20_{}^o = 120_{}^o$
So the values of the three angles are $20_{}^o, 40_{}^o, 120_{}^o$

Note: The perimeter of a triangle is equal to the sum of its three sides
The area of the triangle is always half of the product of its base and height
The side which is opposite to the greatest angle of the triangle is the longest among the three sides of the triangle.
The sum of the interior opposite angles of the triangle is always equal to the exterior angle of the triangle and it is regarded as exterior angle property.
It is to be kept in mind that the sum of the length of two sides of a triangle is always greater than the length of the third side.
In case of all types of triangle, sum of the angles of a triangle is $180_{}^o$
The difference of the length of two sides of a triangle is always less than the third side of the triangle.