(C) Given,$b = 2$,$c = 3$,and $\angle B = \frac{\pi}{6}$.Using the cosine rule in $\triangle ABC$:$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$Substituting t

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(C) Given,$b = 2$,$c = 3$,and $\angle B = \frac{\pi}{6}$.Using the cosine rule in $\triangle ABC$:$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$Substituting the given values:$\cos \frac{\pi}{6} = \frac{a^2 + 3^2 - 2^2}{2 \cdot a \cdot 3}$$\frac{\sqrt{3}}{2} = \frac{a^2 + 9 - 4}{6a}$$\frac{\sqrt{3}}{2} = \frac{a^2 + 5}{6a}$Multiplying both sides by $6a$:$3\sqrt{3} a = a^2 + 5$Rearranging the terms,we get:$a^2 - 3\sqrt{3} a + 5 = 0$

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

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In a $\triangle ABC$,if $b=2, c=3$ and $\angle B=\frac{\pi}{6}$,then $a$ satisfies the equation

AP EAMCET 2021 All AP EAMCET

A
$a^2+3 \sqrt{3} a+5=0$
B
$a^2+3 \sqrt{3} a-5=0$
C
$a^2-3 \sqrt{3} a+5=0$
D
$\sqrt{3} a^2+3 a+5=0$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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In a $\triangle ABC$,if $b=2, c=3$ and $\angle B=\frac{\pi}{6}$,then $a$ satisfies the equation

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In a $\Delta ABC$,$\angle A = 30^\circ$ and $a = 8 \text{ cm}$,then the distance of the orthocentre from vertex $A$ is equal to (with usual notations).

In a $\triangle ABC$,the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ equals:

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