1
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

In a triangle ABC , with usual notations. $\frac{2 \cos \mathrm{~A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{2 \cos \mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}$. Then $\angle \mathrm{A}=$

A
$\frac{\pi}{4}$
B
$\frac{\pi}{6}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{3}$
2
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If in triangle ABC , with usual notations $\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{C}}{2}=\sin \frac{\mathrm{B}}{2}$ and 2 s is the perimeter of the triangle, then the value of $s$ is

A
2 b
B
b
C
$4 b$
D
$\frac{\mathrm{b}}{2}$
3
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{A} \equiv(0,0), \mathrm{B}(3,0), \mathrm{C}(0,-4)$ are vertices of $\triangle A B C$, then the co-ordinates of incentre of $\triangle \mathrm{ABC}$ is

A
$\left(\frac{45}{14}, \frac{3}{14}\right)$
B
$\left(\frac{45}{14}, \frac{-3}{14}\right)$
C
$\left(\frac{3}{14}, \frac{45}{14}\right)$
D
$\left(\frac{-3}{14}, \frac{45}{14}\right)$
4
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0
In a triangle ABC with usual notations if $b \sin C(b \cos C+c \cos B)=42$, then area of triangle $\mathrm{ABC}=$
A
42 sq. units
B
21 sq. units
C
24 sq. units
D
12 sq. units
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