In triangle ABC,if A is acute,C is obtuse,sin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,in $	riangle ABC$,$A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$. First,calculate $\cos A = \sqrt{1 - \sin^

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Given,in $	riangle ABC$,$A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$. First,calculate $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3\sqrt{3}}{14}ight)^2} = \sqrt{1 - \frac{27}{196}} = \sqrt{\frac{169}{196}} = \frac{13}{14}$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting the values: $\frac{13}{14} = \frac{5^2 + c^2 - 3^2}{2 	imes 5 	imes c} = \frac{25 + c^2 - 9}{10c} = \frac{16 + c^2}{10c}$. Cross-multiplying: $130c = 14(16 + c^2) \Rightarrow 130c = 224 + 14c^2$. Rearranging: $14c^2 - 130c + 224 = 0 \Rightarrow 7c^2 - 65c + 112 = 0$. Factoring: $7c^2 - 49c - 16c + 112 = 0$ $\Rightarrow 7c(c - 7) - 16(c - 7) = 0$ $\Rightarrow (7c - 16)(c - 7) = 0$. So,$c = \frac{16}{7}$ or $c = 7$. Since $C$ is obtuse,the side $c$ must be the longest side. Checking the conditions: If $c = \frac{16}{7} \approx 2.28$,then $a=3, b=5, c=2.28$. Here $b$ is the longest side,so $B$ would be obtuse. If $c = 7$,then $a=3, b=5, c=7$. Here $c$ is the longest side,so $C$ is obtuse. Therefore,$c = 7$.

In triangle $ABC$,if $A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$,then $c =$

Similar Questions

In a $\Delta ABC$,if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$,then its area is

In a $\Delta ABC$,the expression $\left( \frac{a^2}{\sin A} + \frac{b^2}{\sin B} + \frac{c^2}{\sin C} \right) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ simplifies to,where $\Delta$ is the area of the triangle.

The perimeter of a $\Delta ABC$ is $6$ times the arithmetic mean of the sines of its angles. If the side $a$ is $1$,then the angle $A$ is

If in a triangle $ABC$,$a, b, c$ are sides and angle $A$ is given,and $c \sin A < a < c$,and $b_1$ and $b_2$ are two possible values of $b$,then:

If in a $\Delta ABC$,$\cos 3A + \cos 3B + \cos 3C = 1$,then one angle must be exactly equal to .......$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module