Question

A triangle is constructed such that one of its side 8 cm, another side is of length 5.5 cm and the angle opposite to the side of length 5.5 cm is 40${}^{\circ }$. 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 $C_2$, as shown in the image below.

Then x${}^{\circ }$ + y${}^{\circ }$ equals

• A. 160${}^{\circ }$
• B. 180${}^{\circ }$
• C. 140${}^{\circ }$
• D. 120${}^{\circ }$

Answer 1

The correct option is B

180${}^{\circ }$

Step 1:
Draw the base line AB of the length 8 cm.

Step 2:
Measure the angle A with your protractor. Make a mark at 40 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 circle centred at B of radius 5.5 cm. Thus, we have the next step.

Step 3:
Draw a circle 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, using protractor, we see that $\angle A{C}_{1}B$ = 110${}^{\circ }$ and $\angle A{C}_{2}B$ = 70${}^{\circ }$.

i.e., x = 110 and y = 70

Then, x${}^{\circ }$ + y${}^{\circ }$ = 110${}^{\circ }$ + 70${}^{\circ }$ = 180${}^{\circ }$