Question

in triangle ABC,if the side are 7,4√3 and √13 cm ,prove that the smallest angle is 30°.

Answer 1

check the attachment

Answer 2

Answer:

The smallest angle is 30°.

Step-by-step explanation:

Given information: In triangle ABC,if the side are 7,4√3 and √13 cm.

To prove: The smallest angle is 30°.

Proof:

Let a=7, b=4√3 and c=√13.

In a triangle, the smallest angle has shortest opposite side.

c=√13 is shortest side.

Cosine formula:

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Substitute the given value in the above formula.

$\cos C=\frac{(7)^2+(4\sqrt{3})^2-(\sqrt{13})^2}{2(7)(4\sqrt{3})}$

$\cos C=\frac{49+48-13}{56\sqrt{3}}$

$\cos C=\frac{84}{56\sqrt{3}}$

$\cos C=\frac{3}{2\sqrt{3}}$

On rationalization.

$\cos C=\frac{3\times \sqrt{3}}{2\sqrt{3}\times \sqrt{3}}$

$\cos C=\frac{3\times \sqrt{3}}{2\times 3}$

$\cos C=\frac{\sqrt{3}}{2}$

$\cos C=\cos 30^{\circ}$

On comparing both sides.

$C=30^{\circ}$

Hence proved.