1
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
2
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

In a triangle with one of the angles $120^{\circ}$, the lengths of the sides form an A.P. If length of the greatest side is 7 m , then the area of the triangle is

A
$\frac{15 \sqrt{3}}{4} \mathrm{~m}^2$
B
$\frac{15 \sqrt{3}}{2} \mathrm{~m}^2$
C
$\frac{15}{2} \mathrm{~m}^2$
D
$\frac{15}{4} \mathrm{~m}^2$
3
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

In $\triangle \mathrm{ABC}$, with usual notations, if $\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}$, then $a^2=$

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

In a triangle ABC with usual notations, $\cot \frac{\mathrm{A}}{2}+\cot \frac{\mathrm{B}}{2}+\cot \frac{\mathrm{C}}{2}=$

A
$\frac{\mathrm{s}^2}{\Delta}$, where $\Delta$ is the area of the triangle ABC .
B
$\frac{s}{\Delta}$, where $\Delta$ is the area of the triangle ABC .
C
$\frac{\Delta}{\mathrm{s}}$, where $\Delta$ is the area of the triangle ABC .
D
$\Delta$, where $\Delta$ is the area of the triangle ABC .
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