Question

$△$ABC is constructed such that AB=AC= 3 cm and $\angle$A = 120${}^{\circ }$. Then the difference of measures of $\angle$A and $\angle$C is equal to

• A. 60${}^{\circ }$
• B. 90${}^{\circ }$
• C. 30${}^{\circ }$
• D. 45${}^{\circ }$

Answer 1

The correct option is B 90${}^{\circ }$

Step 1:
Draw the line segment AB of the length 3 cm.

Step 2:
Measure the angle A with your protractor. Make a mark at 120 degrees and draw a line from A passing through the mark.

Step 3:
Cut an arc of radius 3 cm on the line passing through the point A. The point where this arc cuts the line is the point C.

Step 4:
Draw a line from C to point B to get the required triangle ABC, as shown below.

From the above construction, we see that the measure AB and AC are equal.

So, $\angle$ B = $\angle$ C

Now, $\angle$A + $\angle$B + $\angle$C = 180${}^{\circ }$
$\angle$B + $\angle$C = 180${}^{\circ }$ - 120${}^{\circ }$

Thus, $\angle$ B = $\angle$ C = 30${}^{\circ }$

So,the difference of measure of $\angle$A and $\angle$C = $\angle$A - $\angle$C = 120${}^{\circ }$ - 30${}^{\circ }$ = 90${}^{\circ }$