1
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
Let $P_1$, $P_2$, $P_3$ be the altitudes of a triangle ABC from the vertices A, B, C respectively. If $\triangle$ denotes the area of the triangle and s is the semi-perimeter of the triangle, then $\dfrac{\cos A}{P_1} + \dfrac{\cos B}{P_2} + \dfrac{\cos C}{P_3} =$
A
$R$
B
$\dfrac{1}{R}$
C
$R^2$
D
$\dfrac{1}{R^2}$
2
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
In $\triangle ABC$, with the usual notations, $\angle C = 90^\circ$, then $\sin(A - B)$ is equal to....
A
$\dfrac{a^2 + b^2}{a^2 - b^2}$
B
$\dfrac{a^2 + c^2}{a^2 - c^2}$
C
$\dfrac{b^2 + c^2}{b^2 - c^2}$
D
$\dfrac{a^2 - b^2}{a^2 + b^2}$
3
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
In $\triangle ABC$ with usual notations, if $1 + \tan\left(\dfrac{A}{2}\right)\tan\left(\dfrac{B}{2}\right) = \dfrac{k}{s}$ (where $s$ is the semi-perimeter), then the value of $k$ is...
A
$2$
B
$a + b - c$
C
$a + b$
D
$s - c$
4
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

With usual notation, in a triangle ABC $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then the value of $\cos B$ is equal to

A

$\frac{17}{35}$

B

$\frac{17}{70}$

C

$\frac{19}{35}$

D

$\frac{19}{70}$

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