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.