Question

△ABC is isosceles with AB = AC = 10. ∠BAC = 30◦. 2
Find BC .

Answer 1

Given:

∆ABC is isosceles with AB = AC = 10.

<BAC = 30°

To Find:

BC = ?

Construction:

Draw AD perpendicular to BC.

Solution:

$In \: \triangle ABC , is \: isosceles .$

$AB = AC = 10$

$\angle B = \angle C$

$\blue { ( Angles \: opposite \: to \: equal }$

$\blue { sides \: are \: equal )}$

/* By Angle Sum Property */

$\angle A + \angle B + \angle C = 180\degree$

$\implies 30\degree + \angle B + \angle B = 180\degree$

$\implies 2\angle B = 180\degree - 30 \degree$

$\implies 2\angle B = 150\degree$

$\implies \angle B = 75\degree$

$Now , In \: \triangle {ABD}, AB = 10$

$Cos \angle B = \frac{BD}{AB}$

$\implies cos 75 \degree = \frac{BD}{10}$

$\implies cos ( 45 + 30 ) = \frac{BD}{10}$

$\implies Cos 45Cos30 - sin45sin 30 = \frac{BD}{10}$

$\implies \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{BD}{10}$

$\implies \frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{2}} = \frac{BD}{10}$

$\implies \frac{BD}{10} = \frac{\sqrt{3}-1}{2\sqrt{2}}$

$\implies BD = \frac{10(\sqrt{3}-1)}{2\sqrt{2}}$

$\implies BD = \frac{5(\sqrt{3}-1)}{\sqrt{2}}$

$\red{BC} = 2 \times BD$

$= 2\times \frac{5(\sqrt{3}-1)}{\sqrt{2}}$

$\red{BC}\green{ = \frac{10(\sqrt{3}-1)}{\sqrt{2}}}$

•••♪

Answer 2

Answer:

Step-by-step explanation:

Let in triangle ABC , AB = AC = x (let) , D is the mid point of BC.

In triangle CDA

CD/AC=sin15°

5/AC= sin(45°-30°)=(3^1/2 - 1)/2.2^1/2

AC = x =(10.2^1/2)/(3^1/2 -1)

x=(10.2^1/2) (3^1/2+1)/(3^1/2 -1) (3^1/2 +1)

x=(10.2^1/2) (3^1/2 +1)/2 =5.2^1/2(3^1/2+1)

Area of triangle ABC =(1/2).AB.AC.sin30°

= (1/2).x.x.(1/2)

= (1/4).x^2 =(1/4).50.(3^1/2 + 1)^2 cm^2

=(50/4)(3+1+2.3^1/2)

=(25/2).2.(2+3^1/2)

=25.(2+3^1/2) cm^2. Answer.

Refer the attachment

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