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

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

Question

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

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Given: Area $= 10 \sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,Perimeter $P = a + b + c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} = \frac{1}{2} ac \sin B$.$10 \sqrt{3} = \frac{1}{2} ac \sin 60^{\circ} \implies 10 \sqrt{3} = \frac{1}{2} ac \left( \frac{\sqrt{3}}{2} ight) \implies ac = 40$.Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.$\cos 60^{\circ} = \frac{a^2 + c^2 - b^2}{2(40)} \implies \frac{1}{2} = \frac{(a+c)^2 - 2ac - b^2}{80}$.Since $a+c = 20 - b$,we have $40 = (20-b)^2 - 2(40) - b^2$.$40 = 400 - 40b + b^2 - 80 - b^2$.$40 = 320 - 40b$.$40b = 280 \implies b = 7$.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

Viruses behave as non-living entities because outside the host cell they:

$A$ coil of wire of radius $r$ has $600$ turns and a self-inductance of $108 \ mH$. The self-inductance of a coil with the same radius and $500$ turns is (in $mH$)

$A$ glucose molecule reacts with $X$ number of molecules of phenylhydrazine to yield osazone. The value of $X$ is

Which structure in the embryo is considered the precursor to the urinary bladder?

$A$ tension of $22 ~N$ is applied to a copper wire of cross-sectional area $0.02 ~cm^2$. The Young's modulus of copper is $1.1 \times 10^{11} ~N/m^2$ and the Poisson's ratio is $0.32$. The decrease in cross-sectional area will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module