(C) Using the Law of Cosines,the formula for $\cos A$ is given by: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ Substituting the given values $a = 2$,$b = 3

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$\cos^{-1}\left(\frac{7}{8}\right)$

(C) Using the Law of Cosines,the formula for $\cos A$ is given by: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ Substituting the given values $a = 2$,$b = 3$,and $c = 4$: $\cos A = \frac{3^2 + 4^2 - 2^2}{2(3)(4)}$ $\cos A = \frac{9 + 16 - 4}{24}$ $\cos A = \frac{21}{24}$ $\cos A = \frac{7}{8}$ Therefore,$A = \cos^{-1}\left(\frac{7}{8}\right)$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Difficult

In $\Delta ABC$,$a = 2 \ cm$,$b = 3 \ cm$,and $c = 4 \ cm$,then angle $A$ is

A
$\cos^{-1}\left(\frac{1}{24}\right)$
B
$\cos^{-1}\left(\frac{11}{16}\right)$
C
$\cos^{-1}\left(\frac{7}{8}\right)$
D
$\cos^{-1}\left(-\frac{1}{4}\right)$

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Relation between sides and angles, Solutions of triangles

If in a triangle $ABC$ right-angled at $B$,$s - a = 3$ and $s - c = 2$,then the values of $a$ and $c$ are respectively:

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In triangle $ABC$ with usual notations $b=\sqrt{3}$,$c=1$,and $m \angle A=30^{\circ}$,then the largest angle of the triangle is (in $^{\circ}$)

EasyMHT CET 2023

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If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

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In $\triangle ABC$, with usual notations, if $\cos A = \frac{1}{2}$, $a = 3$, and $\angle B = \frac{\pi}{6}$, then the values of $b$ and $c$ are:

DifficultMHT CET 2026

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In a $\triangle ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to

MediumTS EAMCET 2004

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In $\Delta ABC$,$a = 2 \ cm$,$b = 3 \ cm$,and $c = 4 \ cm$,then angle $A$ is

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If in a triangle $ABC$ right-angled at $B$,$s - a = 3$ and $s - c = 2$,then the values of $a$ and $c$ are respectively:

In triangle $ABC$ with usual notations $b=\sqrt{3}$,$c=1$,and $m \angle A=30^{\circ}$,then the largest angle of the triangle is (in $^{\circ}$)

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