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Math  /  Trigonometry

Questiona) Show that the cosine rule shown below can be rearranged to give cos⁡A=b2+c2−a22bc\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} b) What is the size of angle θ\theta in the triangle below? Give your answer to the nearest degree.
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Studdy Solution

STEP 1

1. The cosine rule is applicable to any triangle.
2. The cosine rule is given by a2=b2+c2−2bccos⁡A a^2 = b^2 + c^2 - 2bc \cos A .
3. The triangle in part (b) has sides of 15 mm, 23 mm, and 12 mm, with θ\theta opposite the 12 mm side.

STEP 2

1. Rearrange the cosine rule to express cos⁡A\cos A.
2. Apply the rearranged cosine rule to find the angle θ\theta.
3. Calculate the angle θ\theta and round to the nearest degree.

STEP 3

Start with the cosine rule:
a2=b2+c2−2bccos⁡A a^2 = b^2 + c^2 - 2bc \cos A
Rearrange to solve for cos⁡A\cos A:
2bccos⁡A=b2+c2−a2 2bc \cos A = b^2 + c^2 - a^2
Divide both sides by 2bc2bc:
cos⁡A=b2+c2−a22bc \cos A = \frac{b^2 + c^2 - a^2}{2bc}

STEP 4

Identify the sides of the triangle for angle θ\theta:
- a=12 a = 12 mm (opposite θ\theta) - b=15 b = 15 mm - c=23 c = 23 mm
Substitute these values into the rearranged cosine rule:
cos⁡θ=152+232−1222×15×23 \cos \theta = \frac{15^2 + 23^2 - 12^2}{2 \times 15 \times 23}

STEP 5

Calculate the cosine of θ\theta:
cos⁡θ=225+529−1442×15×23 \cos \theta = \frac{225 + 529 - 144}{2 \times 15 \times 23} =610690 = \frac{610}{690} ≈0.8841 \approx 0.8841
Find θ\theta by taking the inverse cosine:
θ=cos⁡−1(0.8841) \theta = \cos^{-1}(0.8841)
Calculate θ\theta and round to the nearest degree:
θ≈27∘ \theta \approx 27^\circ
The size of angle θ\theta is:
27∘ \boxed{27^\circ}

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