Two sides of a triangular piece of cardboard are 32 m and 47 m. the angle between these sides is 52°.?

Question

Find the area of the cardboard. use the law of sines or cosines

Answer 1

Given the sides a and b of a triangle and the included angle theta, we have the following area formula:

Area = (1/2)a*b*sin(theta)
=(1/2)*32*47*sin(52 degrees)
= 592.58

Answer 2

All you need is the definition of sine. Take the sine of the angle and multiply it by one of the adjacent sides... that equals the altitude of the triangle with respect to the other adjacent side. Multiply that by the other adjacent side and divide by two.

sin 52° ⋅ 32m ⋅ 47m ⁄ 2 ≈ 592.58…

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If you wanted to take a different approach

Use the Law of Cosines to start
c² = a² + b² − 2·a·b·cos C
c² = (32 m)² + (47 m)² − 2·(32 m)·(47 m)·cos (52°)
c² = 1024 m² + 2209 m² − 3008 m²·cos (52°)
c² ≈ 3233 m² − 3008 m²·(0.615661…)
c² ≈ 3233 m² − 1851.9097… m²
c² ≈ 1381.090… m²
c ≈ √(1381.090… m²)
c ≈ 37.163… m²

Then you could use Herons formula, if you wanted.

Answer 3

A = (1/2) (32) (47) sin 52° m²
A = 593 m ² (to nearest whole number)

Answer 4

Is there such a large cardboard? Well, ita area
= 1/2*32*47*sin(52 deg) m^2
= 0.5*32*47*(0.7880) m^2
= 592.596 m^2

Answer 5

Area = 1/2 * Base * Height
Base = 32 m
Height = 47* sin 52 = 47 * 0.788 = 37.036 m
Area = 0.5 * 32 * 37.036 = 592.576 m^2

Answer 6

Using the law of cosines:

c^2 = a^2 + b^2 - 2abcos(C)
c^2 = 32^2 + 47^2 - 2*32*47*cos(52)
c^2 = 1381.1

c = 37.163

Using Heron's formula:

A = sqrt( 2(ab)^2 + 2(ac)^2 + 2(bc)^2 - a^4 - b^4 - c^4)/4
A = sqrt(5618494.4)/4
A = 592.6