Question:
Draw a triangle ABC with side BC=7cm, angle B =45°, angle A =105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of triangle ABC.
Answer:
Given ∠B = 45, ∠A = 105
Since sum of all interior angles in triangle is equal to 180
=> ∠A + ∠B + ∠C = 180
=> 105 + 45 + ∠C = 180
=> 150 + ∠C = 180
=> ∠C = 180 - 150
=> ∠C = 30
Construction:
1. Draw a triangle ABC having side BC = 7 cm, ∠B = 45, ∠C = 30
2. Now, draw a ray BX making an acute angle with BC on the opposite side of vertex A
3. Now, locate 4 points B1 ,B2 , B3 ,B4 on BX
4. Join B3 X and draw a line through B4 parallel to B3 C intersecting extended BC at C1
5. Now, through C1 , draw a line parallel to AC intersecting extended line segment at C1 .
Now, triangle A1 B1 C1 is the required triangle.
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