MCQ
In a $\triangle A B C, a=5, b=4$ and $\cos (A-B)=\frac{31}{32}$ then side c is equal to
- ✓$6$
- B$7$
- C$9$
- D$8$
✓
Answer
Correct option: A.
$6$
(A) We have,
$\tan \left(\frac{ A - B }{2}\right)=\sqrt{\frac{1-\cos ( A - B )}{1+\cos ( A - B )}}=\sqrt{\frac{1-\left(\frac{31}{32}\right)}{1+\left(\frac{31}{32}\right)}}$
$\Rightarrow \frac{a-b}{a+b} \cot \frac{C}{2}=\frac{1}{\sqrt{63}}$
$\Rightarrow \frac{1}{9} \cot \frac{C}{2}=\frac{1}{\sqrt{63}}$
$\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}$
Now, $\cos C =\frac{1-\tan ^2\left(\frac{ C }{2}\right)}{1+\tan ^2\left(\frac{ C }{2}\right)}$
$\Rightarrow \cos C=\frac{1-\left(\frac{7}{9}\right)}{1+\left(\frac{7}{9}\right)}=\frac{1}{8}$
$\therefore \quad c^2=a^2+b^2-2 a b \cos C$
$\Rightarrow c^2=25+16-40 \times \frac{1}{8}=36 \Rightarrow c=6$
$\tan \left(\frac{ A - B }{2}\right)=\sqrt{\frac{1-\cos ( A - B )}{1+\cos ( A - B )}}=\sqrt{\frac{1-\left(\frac{31}{32}\right)}{1+\left(\frac{31}{32}\right)}}$
$\Rightarrow \frac{a-b}{a+b} \cot \frac{C}{2}=\frac{1}{\sqrt{63}}$
$\Rightarrow \frac{1}{9} \cot \frac{C}{2}=\frac{1}{\sqrt{63}}$
$\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}$
Now, $\cos C =\frac{1-\tan ^2\left(\frac{ C }{2}\right)}{1+\tan ^2\left(\frac{ C }{2}\right)}$
$\Rightarrow \cos C=\frac{1-\left(\frac{7}{9}\right)}{1+\left(\frac{7}{9}\right)}=\frac{1}{8}$
$\therefore \quad c^2=a^2+b^2-2 a b \cos C$
$\Rightarrow c^2=25+16-40 \times \frac{1}{8}=36 \Rightarrow c=6$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
→
The slope of the tangent to the curve $x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:
→If $I(m, n)=\int_0^1 t^m(1-t)^n d t$, then the expression for I(m, n) in terms of $I ( m +1, n -1)$ is
→If equation $\tan \theta+\tan 2 \theta+\tan \theta \tan 2 \theta=1, \theta=$
→The value of $(\hat{i}+\hat{j}) \cdot[(\hat{j}+\hat{k}) \times(\hat{k}+\hat{i})]$ is
→Given that $X \sim B(n, p)$. If $n=10, p=0.4$, then $E(X)=$_____
→If area bounded by the curves $y^2=4 a x$ and $y=m x$ is $\frac{a^2}{3}$, then the value of $m$ is
→The line $\frac{x-2}{3}=\frac{y-3}{4} ; z=4$ is parallel to
→If $1+\cot \theta=\operatorname{cosec} \theta$, then the general value of $\theta$ is
→The angle between the lines $\frac{x-2}{3}=\frac{y+1}{-2}; z=2$ and $\frac{x-1}{2}=\frac{y+\frac{3}{2}}{3}=\frac{z+5}{4}$ is
→