Test Index

AP and TS EAMCET Engineering Model Paper 20

© examsnet.com

Question : 27 of 160

Marks: +1, -0

If in a

\Delta ABC, r_1=2 r_2=3 r_3,

b: c=

4 : 3
5 : 6
4 : 5
5 : 4

Solution:

r_1=2 r_2=3 r_3

\Rightarrow s\tan \frac{A}{2}=2 s\tan \frac{B}{2}=3 s\tan \frac{C}{2}

⇒

\frac{\tan \frac{A}{2}}{6}= \frac{\tan B}{3}= \frac{\tan \frac{C}{2}}{2}= k

\text{since} \frac{A}{2}+ \frac{B}{2}+ \frac{C}{2}=90^{\circ},

We have

\tan \frac{A}{2} \tan \frac{B}{2}+ \tan \frac{B}{2} \tan \frac{C}{2}

+ \tan \frac{C}{2} \tan \frac{A}{2}=1

⇒

(6k)(3k) + (3k)(2k) + (2k)(6k) = 1

⇒

36k^2 = 1

⇒

k=\frac{1}{6}

\sin A= \frac{2 \tan \frac{A}{2}}{1+\tan^{2} \left( \frac{A}{2} \right)}

= \frac{12k}{1+36k^{2}}=1=\frac{5}{5}

and

\sin B = \frac{2 \tan \frac{B}{2}}{1+\tan^{2} \left( \frac{B}{2} \right)}

= \frac{6k}{1+9k^{2}}= \frac{4}{5}

\text{now} \frac{a}{b}= \frac{\sin A}{\sin B}= \frac{5}{4} \Rightarrow a:b=5:4

\sin C= \frac{2 \tan \frac{C}{2}}{1+\tan^{2} \left( \frac{C}{2} \right)}

= \frac{2 \times 2k}{1+4k^{2}}= \frac{4k}{1+4k^{2}}= \frac{3}{5}

\text{Now} \frac{b}{c} = \frac{\frac{4}{5}}{\frac{3}{5}} \Rightarrow \frac{4}{3}

b : c = 4 : 3

© examsnet.com

Go to Question:

Prev Question

Next Question