Question

Draw a triangle ABC with side BC = 6 cm,$\angle C={30}^{\circ }$ and $\angle A={105}^{\circ }$. Then construct another triangle whose sides are times the corresponding $\Delta$ side of ABC.

Answer 1

Given:
$BC=6cm,\angle C={30}^{\circ }$ and $\angle A={105}^{\circ }$
Therefore $\angle B={180}^{\circ }-(\angle A+\angle C)(In\Delta ABC,\angle A+\angle B+\angle C={180}^{\circ })$
$={180}^{\circ }-({105}^{\circ }+{30}^{\circ })={45}^{\circ }$
Steps of construction:
1. draw a line BC =6 cm.
2. draw a ray CN making an angle of ${30}^{\circ }$ at C.
3. draw a ray BM making an angle of ${45}^{\circ }$ at B.
4. locate the point of intersection of rays CN and BM and name it as A.
5. ABC is the triangle whose simialr trinagle is to be drawn.
6. Draw any ray BX making anacute angle with BC on the side opposite to the vertex A.
7. Locate 3 (Greater of 2 and 3 in $\frac{2}{3})$ points $B_1,B_2$ and $B_3$ on BX so that $B{B}_1=B_1{B}_2=B_2{B}_3$.
8. Join $B_{3}C$ and draw a line through $B_2$ (smaller of 2 and 3 in $\frac{2}{3})$ parallel to $B_{3}C$ to intersect BC at C'.
9. Draw a line through C' parallel to the line CA to intersect BA at A'.
10. A 'BC' is the required similar triangle whose sides are $\frac{2}{3}$ times the corresponding sides of $\Delta ABC$