In a triangle ABC,with usual notations,if a=5 | vedclass

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

Question

(A) Using the Napier's Analogy,we have $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$.Given $a=5, b=4$,so $\frac{a-b}

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Using the Napier's Analogy,we have $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$.Given $a=5, b=4$,so $\frac{a-b}{a+b} = \frac{5-4}{5+4} = \frac{1}{9}$.Also,$\cos(A-B) = 2\cos^2\left(\frac{A-B}{2}ight) - 1 = \frac{31}{32}$,which gives $\cos^2\left(\frac{A-B}{2}ight) = \frac{63}{64}$.Thus,$	an^2\left(\frac{A-B}{2}ight) = \sec^2\left(\frac{A-B}{2}ight) - 1 = \frac{64}{63} - 1 = \frac{1}{63}$.So,$	an\left(\frac{A-B}{2}ight) = \frac{1}{\sqrt{63}} = \frac{1}{3\sqrt{7}}$.Substituting this into the Napier's formula: $\frac{1}{3\sqrt{7}} = \frac{1}{9} \cot\left(\frac{C}{2}ight)$,which implies $\cot\left(\frac{C}{2}ight) = \frac{9}{3\sqrt{7}} = \frac{3}{\sqrt{7}}$.Then $	an^2\left(\frac{C}{2}ight) = \frac{7}{9}$,so $\cos C = \frac{1-	an^2(C/2)}{1+	an^2(C/2)} = \frac{1-7/9}{1+7/9} = \frac{2/9}{16/9} = \frac{1}{8}$.Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C = 5^2 + 4^2 - 2(5)(4)(\frac{1}{8}) = 25 + 16 - 5 = 36$.Therefore,$c = 6$.

In a triangle $ABC$,with usual notations,if $a=5, b=4$,and $\cos(A-B)=\frac{31}{32}$,then $c=$

Similar Questions

In a $\Delta ABC$,$a, b, A$ are given and $c_1, c_2$ are two values of the third side $c$. The sum of the areas of the two triangles with sides $(a, b, c_1)$ and $(a, b, c_2)$ is

In a $\triangle ABC$,if $a+3b=3c$,then $\sin \frac{A}{2} =$

In $\triangle ABC$,$(r_1+r_2) \operatorname{cosec}^2 \frac{C}{2} =$

In a $\Delta ABC$,side $b$ is equal to

The angles of a triangle are in the ratio $5:1:6$. The ratio of the smallest side to the greatest side is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module