Sides of a triangle are 2 cm,sqrt{6} cm,and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the sides be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.We know that the smallest angle is opposite to the smallest side.Comparing the valu

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let the sides be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.We know that the smallest angle is opposite to the smallest side.Comparing the values: $a = 2 \approx 2$,$b = \sqrt{6} \approx 2.45$,and $c = \sqrt{3} + 1 \approx 1.732 + 1 = 2.732$.The smallest side is $a = 2$.Using the Law of Cosines for angle $A$: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.$\cos A = \frac{(\sqrt{6})^2 + (\sqrt{3} + 1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3} + 1)} = \frac{6 + (3 + 1 + 2\sqrt{3}) - 4}{2\sqrt{6}(\sqrt{3} + 1)} = \frac{6 + 2\sqrt{3}}{2\sqrt{6}(\sqrt{3} + 1)}$.$\cos A = \frac{2(3 + \sqrt{3})}{2\sqrt{6}(\sqrt{3} + 1)} = \frac{\sqrt{3}(\sqrt{3} + 1)}{\sqrt{6}(\sqrt{3} + 1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.Since $\cos A = \frac{1}{\sqrt{2}}$,we have $A = 45^o$.

Sides of a triangle are $2 \ cm$,$\sqrt{6} \ cm$,and $(\sqrt{3} + 1) \ cm$. The smallest angle of the triangle is .....$^o$

Similar Questions

In $\Delta ABC$,$({b^2} - {c^2})\cot A + ({c^2} - {a^2})\cot B + ({a^2} - {b^2})\cot C = $

In $\triangle ABC$,if $\frac{1}{r_1}, \frac{1}{r_2}$ and $\frac{1}{r_3}$ are in arithmetic progression,then $r_2 : r =$

In $\Delta ABC$,if $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$,then the triangle is:

In $\triangle ABC$,what is the value of $\cos A + \cos B + \cos C$?

The perimeter of a $\Delta ABC$ is $6$ times the arithmetic mean of the sines of its angles. If the side $a$ is $1$,then the angle $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module