1
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a $\triangle A B C,\left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B=$

A

0

B

$2 R^2[\sin 2 A-\sin 2 B]$

C

$\left(b^2-a^2\right) \cot (A+B)$

D

$2 R^2[\tan 2 A-\tan 2 B]$

2
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a $\triangle A B C, \frac{\Delta^2}{a^2+b^2+c^2}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\right)=$

A

0

B

1

C

$\Delta$

D

S

3
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $R: r_1: r=5: 12: 2$, then $r+r_3+r_2-r_1=$

A

$\cos A$

B

$\sin A$

C

$2 r r_1$

D

$2 r_1^2 r$

4
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $12 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}-18 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-24 \hat{\mathbf{k}}$ be the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle A B C$, then the position vector of the incentre of $\triangle A B C$ is

A

$12 \hat{i}-15 \hat{j}-51 \hat{k}$

B

$6 \hat{\mathbf{i}}-\frac{15}{2} \hat{\mathbf{j}}-\frac{51}{2} \hat{\mathbf{k}}$

C

$\frac{4}{3} \hat{\mathbf{i}}-\frac{5}{3} \hat{\mathbf{j}}-17 \hat{\mathbf{k}}$

D

$4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-17 \hat{\mathbf{k}}$

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