TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

If two unbiased dice  are rolled simultaneously until a sum of the number appeared on these dice is either 7 or 11, then the probability that 7 comes before 11, is 

Answer:

Explanation:

Let A  be the events sum appeared on two unbiased dice is 7 and B  be the event sum appeared on two unbiased dice is either 7 or 11.

$\therefore$     P(A)  =$\frac{6}{36}$, P(B) = $\frac{6}{36}+\frac{2}{36}=\frac{2}{9}$

$\therefore$   Required probability,

= $P(A)+P(\overline{B}A)+P(\overline{B}\overline{B}A)+P(\overline{B}\overline{B}\overline{B}A)+.....$

   =$\frac{1}{6}+\frac{7}{9}\times\frac{1}{6}+\left(\frac{7}{9}\right)^{2}\times\frac{1}{6}+$.....

=  $\frac{1}{6}\left(\frac{1}{1-\frac{7}{9}}\right)=\frac{1}{6}\times\frac{9}{2}=\frac{3}{4}$

If a,b and c are three non-collinear points and ka+2b+3c is a point in the plane of a,b, c then k=

Answer:

Explanation:

a,b, and c are three non-collinear point and ka+2b+3c is a point in the plane a,b, c  

$\therefore$     ka+2b+3c=0

 $\Rightarrow$    k+2+3=0 $\Rightarrow$   k=-5

Corresponding  to a triangle ABC, match the items given in List -I  with the items given in List II.

The correct match is 

Answer:

Explanation:

In  $\triangle $ ABC

(A)  We have

      $r r_{2}=r_{1}r_{3}$

$\therefore$    $\frac{\triangle}{s}.\frac{\triangle}{s-b}=\frac{\triangle}{s-a}.\frac{\triangle}{s-c}\Rightarrow\frac{(s-a)(s-c)}{s(s-b)}=1$

$\Rightarrow$      $\tan ^{2} \frac{B}{2}$=1 $\Rightarrow$  $\tan \frac{B}{2}$ = $\tan 45^{0} \Rightarrow$  B=$ 90^{0}$

$\therefore$      $b^{2}  = a^{2}+c^{2}$

 $\therefore$    A→ II

 (B)   We have

 $r_{1}+r_{2}=r_{3}-r$

$\frac{\triangle}{s-a}+\frac{\triangle}{s-b}=\frac{\triangle}{s-c}-\frac{\triangle}{s}$

  $\Rightarrow $     $\frac{s-b+s-a}{(s-a)(s-b)}=\frac{s-s+c}{s(s-c)}$

 $\Rightarrow$    $ \frac{2s-(a+b)}{(s-a)(s-b)}=\frac{c}{s(s-c)}= \tan ^{2} \frac{C}{2}=1$

$\Rightarrow$     $\angle c=90^{0}$

$\Rightarrow$     B  →  III

(C)   $r_{1}=r+2R$

$\frac{\triangle}{s-a}=\frac{\triangle}{s}+\frac{a}{\sin A}\Rightarrow \sin A=\frac{s(s-a)}{\triangle}$

 $\Rightarrow$    $ 2\sin\frac{A}{2}\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\cot \frac{A}{2}$

$\Rightarrow$  $\sin ^{2} \frac{A}{2}=\frac{1}{2}\Rightarrow \sin \frac{A}{2}=\frac{1}{\sqrt{2}}\Rightarrow \angle A=90^{0}$

$\therefore$    C  →  I

If any triangle  ABC. If a:b:c =2:3:4 , then R:r=

Answer:

Explanation:

We have,   

a:b:c =2:3:4 

 Let a=2k,b=3k, c=4k

 $\therefore$     $R= \frac{abc}{4\triangle}$ and   $r=\frac{\triangle}{s}$

 $\therefore$     $\frac{R}{r}= \frac{s.abc}{4\triangle^{2}} $

 $\Rightarrow$      $\frac{R}{r}= \frac{s.(2k)(3k)(4k)}{4s(s-2k)(s-3k)(s-4k)} $

$\Rightarrow$      $\frac{R}{r}= \frac{6k^{3}}{\left(\frac{9k}{2}-2k\right)\left(\frac{9k}{2}-3k\right)\left(\frac{9k}{2}-4k\right)} $

                                                                     $\left[ \because s=\frac{a+b+c}{2}\right]$                                

$\frac{R}{r}= \frac{6.2^{3}.k^{3}}{5k.3k.k} \Rightarrow\frac{R}{r}=\frac{16}{5}$

R:r= 16:5

$\sin ^{2} (3^{0})+\sin ^{2} (6^{0})+\sin ^{2} (9^{0})+....+\sin ^{2} (84^{0})+\sin ^{2} (87^{0})+\sin ^{2} (90^{0})=$

Answer:

Explanation:

We have,

$\sin ^{2} (3^{0})+\sin ^{2} (6^{0})+\sin ^{2} (9^{0})+....+\sin ^{2} (84^{0})+\sin ^{2} (87^{0})+\sin ^{2} (90^{0})$

=$\sin ^{2} (3^{0})+\sin ^{2} (87^{0})+\sin ^{2} (6^{0})+\sin ^{2} (84^{0})+...........\sin ^{2} (90^{0})=$

  =1+1+1+.... 15 times + $\left(\frac{1}{\sqrt{2}}\right)^{2}=15+\frac{1}{2}=\frac{31}{2}$

• Mathematics - General questions