Question

$△$ABC is constructed such that AB = 5.5 cm, BC = 6 cm and $\angle$A = 60${}^{\circ }$. Then the difference between the measure of $\angle$B and $\angle$C is

• A. 7${}^{\circ }$
• B. 12${}^{\circ }$
• C. 14${}^{\circ }$
• D. 17${}^{\circ }$

Answer 1

The correct option is C

14${}^{\circ }$

Step 1:
Draw the base line AB of the length 5.5 cm.

Step 2:
Measure the angle A with your protractor. Make a mark at 60 degrees and draw a line from A passing through the mark.

Now we must locate C. It should be 6 centimetres from B and also should be on the upper line, in order to form the required triangle. All points at a distance of 6 cms from B are on the circle centred at B of radius 6 cm. Thus we have the next step.

Step 3:
Draw a circle with centre at B and radius 6 cm. The point at which this circle cuts the line through A is the required point C.

Thus $△$ABC is formed.

From the above construction, we see that the measure of angle B is 67${}^{\circ }$ and measure of angle C is 53${}^{\circ }$.

Now, measure of angle B - measure of angle C = 67${}^{\circ }$ - 53${}^{\circ }$ = 14${}^{\circ }$