In the triangle ABC, AB = 5 cm, BC = 6 cm and area of triangle is 11.25 cm². What is the value of angle ∠ABC?

Explanation:

Another Explanation (5): ```html Let \(a\), \(b\), and \(c\) be the lengths of sides BC, AC, and AB respectively. Let \(A\) be the area of the triangle. Given: \(c = AB = 5\) cm, \(a = BC = 6\) cm, and \(A = 11.25\) cm². We want to find the angle ∠ABC, which we can denote as \(B\). The area of a triangle can be expressed as: \[A = \frac{1}{2}ac\sin(B)\] Plugging in the given values: \[11.25 = \frac{1}{2} \times 6 \times 5 \times \sin(B)\] \[11.25 = 15 \sin(B)\] \[\sin(B) = \frac{11.25}{15}\] \[\sin(B) = \frac{1125}{1500}\] \[\sin(B) = \frac{225}{300}\] \[\sin(B) = \frac{45}{60}\] \[\sin(B) = \frac{3}{4}\] Therefore, \[B = \sin^{-1}\left(\frac{3}{4}\right)\] So, the angle ∠ABC is \(\sin^{-1}(\frac{3}{4})\). 🎉 ```