Question

A triangle is constructed such that one of its side 10 cm, another side is of length 9 cm and the angle opposite to the side of length 9 cm is 60${}^{\circ }$. 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.

(image here)

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

• A. 78${}^{\circ }$
• B. 82${}^{\circ }$
• C. 86${}^{\circ }$
• D. 89${}^{\circ }$

Answer 1

The correct option is D

89${}^{\circ }$

Step 1:
Draw the base line AB of the length 10 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 9 centimetres from B and also should be on the upper line, in order to form the required triangle. All points at a distance of 9 cms from B are on the circle centred at B of radius 9 cm. Thus we have the next step.

Step 3:
Draw a circle with centre at B and radius 9 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 $C_2$.

(image here)

Now, from the above construction we see that $\angle AB{C}_1$ = 15${}^{\circ }$ and $\angle A{C}_{2}B$ = 74${}^{\circ }$.

i.e., x = 74 and y = 15

Then, x${}^{\circ }$ + y${}^{\circ }$ = 74${}^{\circ }$ + 15${}^{\circ }$ = 89${}^{\circ }$