Question

Given △ABC. m∠A>m∠B>m∠C. Perimeter=30. Which side of △ABC may have length 7?

Answer 1

Solution-

As given in △ABC,

$m\angle A>m\angle B>m\angle C$

As from the properties of trigonometry we know that, the greater the angle is, the greater is the value of its sine. i.e

$\sin A>\sin B>\sin C$

According to the sine law,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

In order to make the ratio same, even though m∠A>m∠B>m∠C, a must be greater than b and b must be greater than c.

$\Rightarrow a>b>c$

Also given that its perimeter is 30. Now we have to find out whose side length is 7. So we have 3 cases.

Case-1. Length of a is 7

As a must be the greatest, so b and c must be less than 7. Which leads to a condition where its perimeter won't be 30. As no 3 numbers less than 7 can add up to 30.

Case-2. Length of b is 7

As b is greater than c, so c must 6 or less than 6. But in this case the formation of triangle is impossible. Because the triangle inequality theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side. If b is 7 and c is 6, then a must be 17. So no 2 numbers below 7 can add up to 17.

Case-3. Length of c is 7

As this is the last case, this must be true.

Therefore, by taking the aid of process of elimination, we can deduce that side c may have length 7.