How can I find the area of a triangle with a given angle and two sides?

[xyz_1965]

Find the area of a triangle with angle 70° in between sides 6 cm and 4 cm.

Solution:

From the SOH-CAH-TOA mnemonic, I want the ratio of the opposite side (CD) to the hypotenuse (AC). I should be using the *sine* function, not cosine. Yes?

SOH leads to sin = opp/hyp

sin(70°) = CD/4

CD = 4 sin(70°)

Here is the rest:

Area = 12 sin(70°)

≈ 12 * 0.9396

Answer:

≈ 11.28 cm^2

Yes?

 

[MarkFL]

I would use:

\(\displaystyle A=\frac{1}{2}ab\sin(C)\)

Using the given data:

\(\displaystyle A=\frac{1}{2}(6\text{ cm})(4\text{ cm})\sin(70^{\circ})\)

Looks good.

 

[xyz_1965]

I would use:

\(\displaystyle A=\frac{1}{2}ab\sin(C)\)

Using the given data:

\(\displaystyle A=\frac{1}{2}(6\text{ cm})(4\text{ cm})\sin(70^{\circ})\)

Looks good.

I know the law of sines or law of cosines could be applied here. This is the next chapter in my studies.