The sides of a triangle are 6+2 sqrt{3},4 sqr | vedclass

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

Question

(A) Given the sides of the triangle are $a = 6+2 \sqrt{3}$,$b = 4 \sqrt{3}$,and $c = \sqrt{24} = 2 \sqrt{6}$.Comparing the values: $a \approx 6 + 3.46

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(A) Given the sides of the triangle are $a = 6+2 \sqrt{3}$,$b = 4 \sqrt{3}$,and $c = \sqrt{24} = 2 \sqrt{6}$.Comparing the values: $a \approx 6 + 3.464 = 9.464$,$b \approx 4 	imes 1.732 = 6.928$,and $c \approx 2 	imes 2.449 = 4.898$.Since $c$ is the smallest side,the smallest angle is $C$.Using the law of cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.$a^2 = (6+2 \sqrt{3})^2 = 36 + 12 + 24 \sqrt{3} = 48 + 24 \sqrt{3}$.$b^2 = (4 \sqrt{3})^2 = 48$.$c^2 = 24$.$\cos C = \frac{48 + 24 \sqrt{3} + 48 - 24}{2(6+2 \sqrt{3})(4 \sqrt{3})} = \frac{72 + 24 \sqrt{3}}{8 \sqrt{3}(3+\sqrt{3})} = \frac{24(3+\sqrt{3})}{8 \sqrt{3}(3+\sqrt{3})} = \frac{3}{\sqrt{3}} = \sqrt{3}$.Wait,checking the calculation again: $\cos C = \frac{72+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})} = \frac{24(3+\sqrt{3})}{16 \sqrt{3}(3+\sqrt{3})} = \frac{24}{16 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$.Thus,$\cos C = \frac{\sqrt{3}}{2}$,which means $C = 30^{\circ}$.The tangent of the smallest angle is $	an 30^{\circ} = \frac{1}{\sqrt{3}}$.

The sides of a triangle are $6+2 \sqrt{3}$,$4 \sqrt{3}$,and $\sqrt{24}$. The tangent of the smallest angle of the triangle is

Similar Questions

In a $\triangle ABC$,if $a=26, b=30$,and $\cos C=\frac{63}{65}$,then $c=$

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 a $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then $a : b =$

In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

Given $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$,then the ordered triple $(\alpha, \beta, \gamma)$ is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module