Question

Draw a triangle ABC with side BC = 6 cm, ∠C = 30° and ∠A = 105°. Then construct another triangle whose sides are 2/3 times the corresponding sides of ∆ABC.​

Answer 1

Step-by-step explanation:

Given:

BC = 6 cm, ∠C = 30° and ∠A = 105°

∴∠B = 180° − (∠A + ∠C) (In ΔABC, ∠A + ∠B + ∠C = 180°)

= 180° − (105° + 30°)

= 45°

Steps of construction:

1. Draw a line BC = 6 cm.

2. Draw a ray CN making an angle of 30° at C.

3. Draw a ray BM making an angle of 45° at B.

4. Locate the point of intersection of rays CN and BM and name it as A.

5. ABC is the triangle whose similar triangle is to be drawn.

6. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

7. Locate 3 (Greater of 2 and 3 in

23) points B1, B2 and B3 on BX so that BB1 = B1B2 = B2B3 .

8. Join B3C and draw a line through B2 (Smaller of 2 and 3 in

23) parallel to B3C 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

23 times the corresponding sides of ΔABC.

Answer 2

Answer:

We know that, sum of angles of triangle is 180⁰.

∴ ∠A + ∠B + ∠C = 180⁰

∴ 105⁰+ 30⁰ + ∠C = 180⁰

∴ 135⁰ + ∠C = 180⁰

∴ ∠C = 45⁰

Step 1 : Draw seg BC of length 6.5 cm

Step 2 : Draw a vector CC’ at an angle of 45⁰ at C.

Step 3 : Draw a vector BB’ at an angle of 30⁰ at B.

Step 4 : Take point of intersection of BB’ and CC’ and label it as A.

2. Construction of ∆PQRtimes corresponding sides of ∆ABC.

Angles of similar triangles are equal.

∴ ∠Q = 30⁰

∴ ∠R = 45⁰

Step 1 : Draw seg QR of length 4.875 cm

Step 2 : Draw a vector RR’ at an angle of 45⁰ at R.

Step 3 : Draw a vector QQ’ at an angle of 30⁰ at Q.

Step 4 : Take point of intersection of QQ’ and RR’ and label it as P.

OR

Let us draw a rough figure,

We know that, tangent is perpendicular to the radius at the point of contact.

∴ ∠D = ∠E = 90⁰

Angle between two tangents is 60⁰

∴ ∠C = 60⁰

Sum of all angles of a quadrilateral is 360⁰.

Therefore, for quadrilateral ADCE,

∴ ∠A + ∠E + ∠C + ∠D = 360⁰

∴ ∠A + 90⁰ + 60⁰ + 90⁰ = 360⁰

∴ ∠A + 240⁰ = 360⁰

∴ ∠A = 120⁰

Construction –

Step 1 : Draw circle with centre A and radius 3 cm.