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# Q.6 Draw a triangle ABC with side BC = 7 cm, $$\angle$$ B = 45°, $$\angle$$ A = 105°. Then, construct a triangle whose sides are $$\frac{4}{3}$$times the corresponding sides of $$\triangle$$ABC.

To find $$\angle$$C:
Given:
$$\angle$$B = 45°, $$\angle$$A = 105°
We know that,
Sum of all interior angles in a triangle is 180°.
$$\angle A+\angle B +\angle C = 180°$$
105°+45°+$$\angle$$C = 180°
$$\angle$$C = 180° ? 150°
$$\angle$$ C = 30°
So, from the property of triangle, we get $$\angle$$C = 30° The required triangle can be drawn as follows.
1. Draw a $$\triangle$$ABC with side measures of base BC = 7 cm, $$\angle$$B = 45°, and $$\angle$$C = 30°.
2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.
4. Join the points B3C.
5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.
6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.
7. Therefore, $$\triangle$$A’BC’ is the required triangle.