Question

$△$ABC is constructed such that AB = 4 cm, BC = 5 cm and $\angle$A = 70${}^{\circ }$. Taking side BC as base, another triangle BCD is constructed such that length of side BD is 3 cm and length of side CD is 3 cm. Then $\angle$DBC + $\angle$DCB equals

• A. 58${}^{\circ }$
• B. 68${}^{\circ }$
• C. 78${}^{\circ }$
• D. 88${}^{\circ }$

Answer 1

The correct option is B

68${}^{\circ }$

Step 1:
Draw the base line AB of the length 4 cm.

Step 2:
Measure the angle A with your protractor. Make a mark at 70 degrees and draw a line from A passing through the mark.

Now we must locate C. It should be 5 centimetres from B and also should be on the upper line, in order to form the required triangle. All points at a distance of 5 cms from B are on the circle centred at B of radius 5 cm. Thus we have the next step.

Step 3:
Draw a circle with centre at B and radius 5 cm. The point at which this circle cuts the line through A is the required point C.

Thus $△$ABC is formed.

Taking side BC as the base, we shall construct triangle BCD.

Step 4:
Use your compass to make a circle with centre at point B and the radius 3 cm.

Step 5:
Use your compass to make a circle with centre at point C and the radius 3 cm.

Step 6:
In the point of intersection between the two circles, you'll find the point D.
Complete the triangle by drawing the lines between all three points.

Now, by above construction, we see that $\angle$DCB = 34${}^{\circ }$ = $\angle$DBC

Therefore $\angle$DCB + $\angle$DBC = 34${}^{\circ }$ + 34${}^{\circ }$ = 68${}^{\circ }$