Question

In a $\Delta ABC$, A:B:C=3:5:4 then $[a+b+c\sqrt{2}]$
A. 2b
B. 2c
C. 3b
D. 3a

Answer 1

Hint:
Here we have given a triangle with the ratio of its three sides. In order to determine the value for [a+b+c\sqrt{2}] we have to first solve the ratio using the property of a triangle and calculate the values of sine of angles. Apply the sine rules and simplify the expression. Now, compare the obtained result with the given option.

Formula Used:
Laws of Sines for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$\dfrac{a}{\sin A}= \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\\
\dfrac{\sin A}{a} = \dfrac{\sin B}{b} =\dfrac{\sin C}{c}$

Complete step-by-step solution:
Given a triangle $ABC$ with $A:B:C=3:5:4$
Therefore,
$3x+5x+4x=180^{o}\\
\rightarrow 12x=180^{o}\\
\rightarrow x=\dfrac{180^{o}}{12}\\
\rightarrow x=15^{o}$
Now we have,
$A=3 \times 15^{o}=45^{o}\\
B=5 \times 15^{o}=75^{o} \\
C=4 \times 15^{o}=60^{o}$
Let [a+b+c\sqrt{2}]...(i)
Using Laws of sines we get;
$\dfrac{a}{\sin A}= \dfrac{b}{\sin B} = \dfrac{c}{\sin C}
\Rightarrow \dfrac{a}{\sin 45^{o}}=\dfrac{b}{\sin 75^{o}}=\dfrac{c}{\sin 60^{o}}\\
\Rightarrow \dfrac{a}{\dfrac{1}{\sqrt{2}}}=\dfrac{b}{\dfrac{\sqrt{3}+1}{2\sqrt{2}}}=\dfrac{c}{\dfrac{\sqrt{3}}{2}}\\
\Rightarrow a\sqrt{2}=\dfrac{2\sqrt{2}b}{1+\sqrt{3}}=\dfrac{2c}{\sqrt{3}}...(i)$
From equation (ii) we can write;
$a=\dfrac{2b}{1+\sqrt{3}}\\
c=\dfrac{\sqrt{2}b\sqrt{3}}{1+\sqrt{3}}$
Put these values in the given equation (i) we get;
$[a+b+c\sqrt{2}]=\dfrac{2b}{1+\sqrt{3}}+b+\dfrac{\sqrt{2}b\sqrt{3}}{1+\sqrt{3}}$
On simplification we get;
$[a+b+c\sqrt{2}]=3b$

So, option C is correct.

Note:
This type of question requires knowledge of trigonometric tables. We must know the correct value of sine at different angles. Remember we can find the value of $\sin 75^{o}$ by breaking it into $\sin(45+30)$. You can use any form of representation of sine laws.