Question

- in a right angled triangle ABC, right angle at C, ∆B = 60° & AB= 15 find BC & CA​

Answer 1

Answer:

Steps of construction: 1. Draw a line segment BC = 8 cm. 2. Make a right angle at the point B i.e., ∠CBX = 90° 3. Draw a are of radius 6 cm as centre B which intersect BX at the point A. 4. Join AC  so, ABC is required right angle triangle. 5. Draw a arc taking centre B which intersect AC at the point K and L respectively taking K and L centre draw two arcs of same radius which intersect at the point M. 6. Join BM so, ∠BDC = 90° 7. Draw perpendicular bisector of BC. 8. Draw a circle taking radius equal to OB and centre O which passes through B, D and C. 9. Draw a arc taking centre A and radius equal to AB intersect the circle at point P so, AP is the magnets

Answer 2

Step-by-step explanation:

Solution :

∆ABC is right angled at C, ∠B = 60°, ∠C = 90°

We know that,

In ∆ABC, ∠A + ∠B + ∠C = 180°

∠A + 60° + 90° = 180°

∠A + 150° = 180°

∠A = 180° - 150°

∠A = 30°

We have,

sin A = CB/AB

sin 30° = CB/15

★ ½ × 15 = CB

★ CB = 7.5 units.

From Pythagoras theorem,

AB² = AC² + BC²

15² = AC² + (15/2)²

225 = AC² + (225/4)

AC² = 225/4 - 225

AC² = 225¾

AC = 15√3/2 units.