Question

A triangle is constructed such that one of its side 6 cm, another side is of length 5.5 cm and the angle opposite to the side of length 5.5 cm is 67${}^{\circ }$. This construction results in the formation of two triangles. Let us call the third vertex of the smaller triangle as $C_1$ and the third vertex of the bigger triangle as $C_2$.

Then $\angle A{C}_{1}B$ +$\angle A{C}_{2}B$ equals

• A. 146${}^{\circ }$
• B. 126${}^{\circ }$
• C. 90${}^{\circ }$
• D. 180${}^{\circ }$

Answer 1

The correct option is A 146${}^{\circ }$

Step 1:
Draw the base line AB of the length 6 cm.

Step 2:
Measure the angle A with your protractor. Make a mark at 67 degrees and draw a line from A passing through the mark.

Now, we must locate C. It should be 5.5 cm from B and also should be on the upper line, in order to form the required triangle. All points at a distance of 5.5 cm from B are on the arc centred at B of radius 5.5 cm. Thus, we have the next step.

Step 3:
Draw an arc with centre at B and radius 5.5 cm. This circle cuts the line through A at two points.

Name the third vertex of the smaller triangle as $C_1$ and the third vertex of the bigger triangle as $C_2$.

Now, from the above construction we see that
$\angle A{C}_{1}B={93}^{\circ }$
and
$\angle A{C}_{2}B={53}^{\circ }$.

Then,
$\angle A{C}_{1}B+\angle A{C}_{2}B$
$={93}^{\circ }+{53}^{\circ }$
$={146}^{\circ }$