Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ABC.
Given: A triangle ABC, with BC = 7 cm, ∠B = 45° and ∠A = 105°
We can find the third angle, ∠C.
Since, sum of all angles in a triangle is always equals to 180°, we can write
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180° [∵ ∠A = 105° & ∠B = 45°]
⇒ 150° + ∠C = 180°
⇒ ∠C = 180° - 150°
⇒ ∠C = 30°
Construction – There are systematic steps involved in construction of the ∆ABC.
For 45° construction, we need to construct 90° first. 90° is required because half of 90° is 45°. So if we are able to construct 90°, then it will be easy to construct 45° due to the relationship between the two angles. So, these steps are to be followed:
1. Draw line BC = 7 cm.
2. Using compass, by taking B as center and with any radius, draw an arc intersecting BC at O.
3. Now take O as center and with the same radius of the compass, draw an arc intersecting the arc just drawn at P. Make sure to draw these arcs with the same radius as set in the beginning.
4. Now take P as a center, and again with the same radius draw an arc intersecting the arc drawn I the beginning.
5. Now take P and Q as centers and set the compass at a radius more than 1/2 PQ, and draw arcs from each point P & Q consecutively intersecting each other at R.
6. Draw a line through arc R meeting at B, such that we have a straight line and also ∠CBR = 90°.
7. Now take B as center and set the compass at any radius, and draw an arc such that the arc intersects at line BC and BR at H and I respectively.
8. Similarly again, take H and I as centers and set the compass at a radius more than 1/2 HI and draw an arc from each point H & I such that the arc intersect each other at a point say, J.
9. Meet this point J to B by a straight line JB. This angle makes 45° with line BC, i.e., ∠CBJ = 45°.
Now for 30° construction, 60° need to be constructed first for ease of construction as half of 60° is 30°. So, these steps are to be followed:
1. Taking C as center and setting any radius, draw an arc intersecting BC at O’.
2. Taking O’ as center this time and with same radius, draw an arc at the previous arc intersecting at say, P’.
3. Know that we have got 60° angle, ∠P’CO’ = 60°.
4. Now taking O’ and P’ as centers and setting the compass at a radius more than 1/2 O’P’, draw an arc from each points separately intersecting at a single point J’. (as done in the construction of 45°)
5. Draw a straight line through J’C.
6. Clearly, ∠J’CB = 30°.
After we have constructed the two angles, we need to extend the lines BJ and CJ’ to meet at a point say, A.
Hence, we have constructed the required triangle ABC.
Now, we need to construct another triangle whose sides are 4/3 times of sides of triangle ABC. The steps are:
1. We have already got ∆ABC, so on the same figure draw a line BX strictly at an acute angle with BC but opposite to the side at where vertex A is located.
2. As the sides of new triangle should be 4/3 times sides of the previous triangle, consider the numbers 4 and 3, out of which 4 is bigger and so locate 4 points (say B1, B2, B3, B4) on line BX.
3. Join B3 to C by a straight line B3C.
4. Then join B4 to C’ but make sure B4C’ is parallel to B3C. Also, extend the line BC to BC’.
5. In the end, join A’ to C’ such that A’C’ is parallel to AC.
6. And so, we have got our new triangle A’BC’, which is 4/3 times of ∆ABC.
