The area of the triangle ABC is 10sqrt{3} tex | vedclass

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

Question

(D) Given: $	ext{Area} = 10\sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,and $a+b+c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} = \frac{1}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Given: $	ext{Area} = 10\sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,and $a+b+c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} = \frac{1}{2}ac \sin B$.$10\sqrt{3} = \frac{1}{2}ac \sin 60^{\circ} \Rightarrow 10\sqrt{3} = \frac{1}{2}ac \left(\frac{\sqrt{3}}{2}ight)$.$10\sqrt{3} = \frac{ac\sqrt{3}}{4} \Rightarrow ac = 40$.Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.$b^2 = (a+c)^2 - 2ac - 2ac \cos 60^{\circ}$.Since $a+c = 20-b$,we have $b^2 = (20-b)^2 - 2(40) - 2(40)(0.5)$.$b^2 = 400 + b^2 - 40b - 80 - 40$.$0 = 280 - 40b$.$40b = 280 \Rightarrow b = 7 	ext{ cm}$.Thus,$\ell(AC) = b = 7 	ext{ cm}$.

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

Similar Questions

In a $\Delta ABC$,if $2s = a + b + c$ and $(s - b)(s - c) = x \sin^2 \frac{A}{2}$,then $x =$

If the sides of a triangle are in $A.P.$,then the cotangents of its half-angles will be in

Points $D$ and $E$ are taken on the side $BC$ of a triangle $ABC$ such that $BD = DE = EC$. If $\angle BAD = x$,$\angle DAE = y$,and $\angle EAC = z$,then the value of $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z}$ is:

In a triangle $ABC$,$a = 5, b = 7$ and $\sin A = \frac{3}{4}$,how many such triangles are possible?

The angles $A, B$ and $C$ of a triangle $ABC$ are in $AP$. If $b: c = \sqrt{3}: \sqrt{2}$,then the angle $A$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module