Question

In a $\vartriangle ABC$, $A:B:C = 3:5:4$, then $a + b + c\sqrt 2 $ is equal to
A. $2b$
B. $2c$
C. $3b$
D. $3a$

Answer 1

Hint: Here we have been given the ratio of three angles of a triangle. We will first find the angles of a triangle by using the property of the triangle and then we will use the sine law of the triangle to find the relation between the sides of the triangle and then we will use these relations to get the required value of the expression.

Formula used:
According to sine law;
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$

Complete step by step solution:
It is given that in $\vartriangle ABC$
$A:B:C = 3:5:4$
Now, we will use these ratios between the angles.
Let the $\angle A = 3x$, $\angle B = 5x$ and $\angle C = 4x$
We know that the sum of all the angles of any triangle is always equal to $180^\circ $.
$ 3x + 5x + 4x = 180^\circ $
On adding the terms, we get
$ \Rightarrow 12x = 180^\circ $
Now, we will divide both by 12.
$ \Rightarrow \dfrac{{12x}}{{12}} = \dfrac{{180^\circ }}{{12}} \\
   \Rightarrow x = 15^\circ \\ $
Therefore,
$ \Rightarrow \angle A = 3 \times 15^\circ = 45^\circ $
$ \Rightarrow \angle B = 5 \times 15^\circ = 75^\circ $
$ \Rightarrow \angle C = 4 \times 15^\circ = 60^\circ $
Now, we will use the sine law here.
According to sine law;
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$
Now, we will substitute the value of angles.
$ \Rightarrow \dfrac{a}{{\sin 45^\circ }} = \dfrac{b}{{\sin 75^\circ }} = \dfrac{c}{{\sin 60^\circ }}$
We know that the value of
$ \sin 45^\circ = \dfrac{1}{{\sqrt 2 }} \\
  \sin 75^\circ = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }} \\
  \sin 60^\circ = \dfrac{{\sqrt 3 }}{2} \\ $
Now, we will substitute the values here.
$ \Rightarrow \dfrac{a}{{\dfrac{1}{{\sqrt 2 }}}} = \dfrac{b}{{\dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}}} = \dfrac{c}{{\dfrac{{\sqrt 3 }}{2}}}$
On further simplification, we get
$ \Rightarrow a\sqrt 2 = \dfrac{{2\sqrt 2 b}}{{1 + \sqrt 3 }} = \dfrac{{2c}}{{\sqrt 3 }}$
We will first consider the first two terms.
$ a\sqrt 2 = \dfrac{{2\sqrt 2 b}}{{1 + \sqrt 3 }}$
$ \Rightarrow a = \dfrac{{2b}}{{1 + \sqrt 3 }}$ ……………. $\left( 1 \right)$
Now, we will consider the last two terms.
$ \Rightarrow \dfrac{{2\sqrt 2 b}}{{1 + \sqrt 3 }} = \dfrac{{2c}}{{\sqrt 3 }}$
On cross multiplying the terms, we get
$ \Rightarrow c = \dfrac{{\sqrt 2 b\sqrt 3 }}{{1 + \sqrt 3 }}$ …………….. $\left( 2 \right)$
But here, we have to find the value of the given expression i.e. $a + b + c\sqrt 2 $.
We will substitute the values from equation 1 and equation 2 here.
$ a + b + c\sqrt 2 = \dfrac{{2b}}{{1 + \sqrt 3 }} + b + \sqrt 2 \times \dfrac{{\sqrt 2 b\sqrt 3 }}{{1 + \sqrt 3 }}$
On further simplifying the terms, we get
$\Rightarrow$ $ a + b + c\sqrt 2 = \dfrac{{2b}}{{1 + \sqrt 3 }} + b + \dfrac{{2\sqrt 3 b}}{{1 + \sqrt 3 }}$
On adding the terms, we get
$\Rightarrow$ $a + b + c\sqrt 2 = \dfrac{{2b + \left( {1 + \sqrt 3 } \right)b + 2\sqrt 3 b}}{{1 + \sqrt 3 }}$
On using the distributive property of multiplication, we get
$\Rightarrow$ $a + b + c\sqrt 2 = \dfrac{{2b + b + \sqrt 3 b + 2\sqrt 3 b}}{{1 + \sqrt 3 }}$
On adding all the terms present in the numerator, we get
$\Rightarrow$ $a + b + c\sqrt 2 = \dfrac{{\left( {3 + 3\sqrt 3 } \right)b}}{{1 + \sqrt 3 }}$
On further simplification, we get
$\Rightarrow$ $a + b + c\sqrt 2 = \dfrac{{3\left( {1 + \sqrt 3 } \right)b}}{{1 + \sqrt 3 }} = 3b$

Hence, the correct option is option C.

Note: Here we have obtained the value of the given expression. Here we have used the distributive property of multiplication. According to the distributive property of multiplication, if $a$, $b$ and $c$ are three real numbers then $a \cdot \left( {b + c} \right)$.