(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ Given $a = 4$,$b = 3$,and $A = 60^\circ$,we have: $\cos 60^\circ = \frac{3^2 + c^

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

(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ Given $a = 4$,$b = 3$,and $A = 60^\circ$,we have: $\cos 60^\circ = \frac{3^2 + c^2 - 4^2}{2(3)(c)}$ $\frac{1}{2} = \frac{9 + c^2 - 16}{6c}$ $\frac{1}{2} = \frac{c^2 - 7}{6c}$ $6c = 2(c^2 - 7)$ $6c = 2c^2 - 14$ $2c^2 - 6c - 14 = 0$ Dividing by $2$,we get: $c^2 - 3c - 7 = 0$.

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

Difficult

In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

A
$c^2 - 3c - 7 = 0$
B
$c^2 + 3c + 7 = 0$
C
$c^2 - 3c + 7 = 0$
D
$c^2 + 3c - 7 = 0$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Trigonometrical Equations

Subtopic

Relation between sides and angles, Solutions of triangles

If the sides of a $\Delta$ are $(x^2 + x + 1)$,$(2x + 1)$,and $(x^2 - 1)$,then the greatest angle is.....$^o$

Easy

View Solution

In $\triangle ABC$,if $a = 2b$ and $|A - B| = \frac{\pi}{3}$,then $\angle C = $

DifficultAP EAMCET 2017

View Solution

The smallest angle of the $\Delta ABC$,when $a = 7, b = 4\sqrt{3}$ and $c = \sqrt{13}$,is .....$^o$

Easy

View Solution

If $H$ is the orthocentre of $\triangle ABC$ and $AH=x, BH=y, CH=z$,then $\frac{abc}{xyz}=$

DifficultAP EAMCET 2024

View Solution

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

EasyKCET 2013

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

Similar Questions

If the sides of a $\Delta$ are $(x^2 + x + 1)$,$(2x + 1)$,and $(x^2 - 1)$,then the greatest angle is.....$^o$

In $\triangle ABC$,if $a = 2b$ and $|A - B| = \frac{\pi}{3}$,then $\angle C = $

The smallest angle of the $\Delta ABC$,when $a = 7, b = 4\sqrt{3}$ and $c = \sqrt{13}$,is .....$^o$

If $H$ is the orthocentre of $\triangle ABC$ and $AH=x, BH=y, CH=z$,then $\frac{abc}{xyz}=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module