Question

Use figure shown above. In the given triangle ABC, AC = BC. Find AB and $\angle$ACB.

• A. 2 cm, 60 degrees
• B. 1 cm, 30 degrees
• C. 3 cm, 120 degrees
• D. 2 cm, 120 degrees

Answer 1

The correct option is A

2 cm, 60 degrees

Let us consider triangle BCD. It is a right triangle right angled at D. BC is hypotenuse. So,

${BC}^2$ =${BD}^2$ + ${CD}^2$

$2^2$ = ${BD}^2$ + (${\sqrt{3}}^2$)

4 = ${BD}^2$ + 3

BD = 1 cm

Since AC = BC if CD is perpendicular to AB, it will also bisect AB and also bisect $\angle$ACB.

So, AD = BD AD = 1cm

AB = AD + BD AB = 1 + 1 = 2 cm AB = 2 cm

In right triangle BCD, cos ($\angle$BCD) = $\frac{\sqrt{3}}{2}$

So, $\angle$BCD = ${30}^{\circ }$.

$\angle$ACD = $\angle$BCD

$\angle$ACB = $\angle$ACD + $\angle$BCD

$\angle$ACB = ${30}^{\circ }$ + ${30}^{\circ }$

$\angle$ = ${60}^{\circ }$