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

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To find the area of a triangle with a 70° angle between sides measuring 6 cm and 4 cm, the formula A = 1/2 * a * b * sin(C) is used. Substituting the values, the area is calculated as A = 1/2 * (6 cm) * (4 cm) * sin(70°). This results in an area of approximately 11.28 cm². The discussion also mentions the applicability of the law of sines and law of cosines for further calculations. Understanding these concepts is essential for solving triangle-related problems effectively.
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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?
 
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I would use:

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

Using the given data:

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

Looks good.
 
MarkFL said:
I would use:

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

Using the given data:

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

Looks good.

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