Question

In ΔWXY, w = 4.8 cm, x = 1.5 cm and ∠Y=52°. Find the area of ΔWXY, to the nearest 10th of a square centimeter.

Answer 1

Answer:

Area = 3.0 square cm

Step-by-step explanation:

By cosine formula,

WX² = WY² + XY² - 2(WY)(XY)cos(∠W)

WX² = (1.5)² + (4.8)² - 2(1.5)(4.8)cos(52°)

WX² = 2.25 + 23.04 - 8.866

WX = √(16.424)

WX = 4.05 cm

Since, area of a triangle with known sides = $\sqrt{s(s-a)(s-b)(s-c)}$

Here, s = $\frac{a+b+c}{2}$

And a, b, c are the sides of the triangle

By this formula,

s = $\frac{1.5+4.8+4.05}{2}$

s = 5.2

Therefore, area of the triangle XYZ = $\sqrt{5.2(5.2-1.5)(5.2-4.8)(5.2-4.05)}$

= $\sqrt{5.2(3.7)(0.4)(1.15)}$

= $\sqrt{8.85}$

= 2.97

≈ 3.0 square cm