NIMCET 2027 Properties Of Triangle Previous Year Questions (PYQs) – Page 2 of 3
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Showing 24 questions
If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP, then sin A, sin B, sin C are in
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Solution
If altitudes of a triangle are in HP then its side will be in AP because sides are inverse proportion to height as area is constant. a, b, c are sides of triangle.
a, b, c are in A.P.
sin A, sin B, sin C are in A.P.
In a triangle, if the sum of two sides is x and their product is y such that (x+z)(x-z)=y, where z is the third side of the triangle , then triangle is
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Solution
In a triangle ABC, $a\cos ^2\frac{C}{2}+\, c\, \, {\cos }^2\frac{A}{2}=\frac{3b}{2}$ then the sides of the triangle are in
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Solution
In a triangle ABC, let angle C = π/2. If R is the inradius and R is circumradius of the triangle ABC,
then 2(r + R) equals
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Solution
In $\triangle ABC$, $R$ is circumradius and $8R^2=a^2+b^2+c^2$.
Then $\triangle ABC$ is:
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Solution
In any triangle:
$a^2+b^2+c^2 = 8R^2$ ⇔ right-angled triangle identity.
The value of $\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)+\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$ is:
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Solution
We know that the principal range of $\cos^{-1}x$ is $[0,\pi]$.
Now,
$\cos\left(-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$
So,
$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)=\frac{\pi}{6}$
Also,
$\sin\frac{5\pi}{6}=\frac{1}{2}$
The principal range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.
So,
$\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\sin^{-1}\left(\frac{1}{2}\right)$
$=\frac{\pi}{6}$
Therefore,
$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)+\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$
$=\frac{\pi}{6}+\frac{\pi}{6}$
$=\frac{\pi}{3}$
In a triangle ABC, angle A=90° and D is the midpoint of AC. What is the value of 
equal to?
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Solution
What is the largest area of an isosceles triangle with two edges of length 3?
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Solution
In a triangle $ABC$, $a = 4$, $b = 3$, $\angle BAC = 60^\circ$, then the equation for which $c$ is the root is
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Solution
If A, B and C is three angles of a ΔABC, whose area is Δ. Let a, b and c be the sides opposite to the
angles A, B and C respectively. Is $s=\frac{a+b+c}{2}=6$, then the product $\frac{1}{3} s^{2} (s-a)(s-b)(s-c)$ is equal to
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Solution
NIMCET 2027
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NIMCET 2027
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and More.
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