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.


Answer :

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.

NCERT solutions of related questions for Constructions

NCERT solutions of related chapters class 10 maths

NCERT solutions of related chapters class 10 science