Question

in $delta abc$, $b = 40$ cm, $c = 60$ cm and $angle a = 22^{circ}$. find the area of $delta abc$, to the nearest square centimeter.

Explanation:

Step1: Recall the area - formula for a triangle

The area formula for a triangle with two - side lengths \(b\), \(c\) and the included angle \(A\) is \(S=\frac{1}{2}bc\sin A\).

Step2: Substitute the given values

Given \(b = 40\mathrm{cm}\), \(c = 60\mathrm{cm}\), and \(A = 22^{\circ}\). We know that \(\sin22^{\circ}\approx0.3746\). Substitute these values into the formula: \(S=\frac{1}{2}\times40\times60\times\sin22^{\circ}\).

Step3: Calculate the area

First, calculate \(\frac{1}{2}\times40\times60 = 1200\). Then, \(S = 1200\times0.3746=449.52\mathrm{cm}^{2}\).

Step4: Round the result

Rounding \(449.52\) to the nearest square - centimeter gives \(450\mathrm{cm}^{2}\).

Answer:

\(450\mathrm{cm}^{2}\)