‘Find the value of x in the triangle shown below Find the value of & in the triangle shown below: 1 5.7 x 6 6 56’
The Correct Answer and Explanation is:
To solve for xxx in the triangle, we must assume the triangle involves sides and angles in a known configuration. Based on your input — “1 5.7 x 6 6 56” — it seems we are dealing with a triangle with:
- One angle measuring 56∘56^\circ56∘
- Two known sides, likely: one of length 5.7, another of length 6
- The side xxx is opposite the 56∘56^\circ56∘ angle.
We can apply the Law of Cosines or Law of Sines, depending on the given arrangement.
Let’s assume this configuration:
- Side a=xa = xa=x is opposite the known angle A=56∘A = 56^\circA=56∘
- Side b=5.7b = 5.7b=5.7
- Side c=6c = 6c=6
We will apply the Law of Cosines if we are given two sides and the included angle:a2=b2+c2−2bccosAa^2 = b^2 + c^2 – 2bc \cos Aa2=b2+c2−2bccosA
Substituting the values:x2=5.72+62−2(5.7)(6)cos(56∘)x^2 = 5.7^2 + 6^2 – 2(5.7)(6)\cos(56^\circ)x2=5.72+62−2(5.7)(6)cos(56∘)
First calculate:
- 5.72=32.495.7^2 = 32.495.72=32.49
- 62=366^2 = 3662=36
- 2⋅5.7⋅6=68.42 \cdot 5.7 \cdot 6 = 68.42⋅5.7⋅6=68.4
- cos(56∘)≈0.5592\cos(56^\circ) \approx 0.5592cos(56∘)≈0.5592
Now plug in:x2=32.49+36−68.4⋅0.5592x^2 = 32.49 + 36 – 68.4 \cdot 0.5592×2=32.49+36−68.4⋅0.5592×2=68.49−38.26=30.23x^2 = 68.49 – 38.26 = 30.23×2=68.49−38.26=30.23x=30.23≈5.5x = \sqrt{30.23} \approx 5.5x=30.23≈5.5
Final Answer:
x≈5.5x \approx \boxed{5.5}x≈5.5
Explanation:
To solve for an unknown side in a triangle when two sides and the included angle are known, we use the Law of Cosines. This rule allows us to relate all three sides and the angle between them. We squared the known side lengths, subtracted the product involving the cosine of the angle, and took the square root of the result. The cosine function helps adjust for the angle’s influence on the length of the opposite side. After careful calculation, we found that the side opposite the 56-degree angle is about 5.5 units long.
