Question

Construct a $\Delta ABC$, whose perimeter is $10.5cm$ and base angles are ${60}^o$ and ${45}^o$. Find the third angle.

• A. ${75}^o$
• B. ${45}^o$
• C. ${90}^o$
• D. ${60}^o$

Answer 1

The correct option is A ${75}^o$

1. Draw a line segment PQ such that AB + BC + AC = PQ.
2. Construct $\angle XPQ=\angle B={60}^{0}and\angle YQP=\angle C={45}^0$
3. Bisect $\angle XPQand\angle YQP$ and their bisectors will meet at A.
4. Draw the perpendicular bisector DE of AP which meets PQ at B.
5. Draw the perpendicular bisector FG of AQ which meets PQ at C.
6. Join AB & AC.
7. ABC is the required triangle.
Sum of interior angle of triangle= ${180}^o$

so third angle = ${180}^o-{45}^o-{60}^o={75}^o$