IB Mathematics SL 3.3 Applications of right and non-right angled trigonometry AI HL Paper 1- Exam Style Questions- New Syllabus

(a)

Let \(\theta=\widehat{BAC}\). The side opposite \(\theta\) is \(BC=8\text{ cm}\), so we use the cosine rule:

\(8^2=15^2+10^2-2(15)(10)\cos\theta\).

Rearranging,

\(\cos\theta=\dfrac{15^2+10^2-8^2}{2(15)(10)}\).

\(\cos\theta=\dfrac{225+100-64}{300}=\dfrac{261}{300}=0.87\).

\(\theta=\cos^{-1}(0.87)=29.5413\ldots^\circ\).

✅ Answer: \(\widehat{BAC}=29.5^\circ\)

(b)

The shaded area is the area of triangle \(ABC\) minus the area of sector \(ABD\).

Area of triangle \(ABC\):

\(\text{Area}=\dfrac{1}{2}(15)(10)\sin(29.5413\ldots^\circ)\).

\(\text{Area of triangle }ABC=36.9788\ldots\text{ cm}^2\).

Since the arc has centre \(A\), both \(AB\) and \(AD\) are radii. Therefore, the radius of sector \(ABD\) is \(10\text{ cm}\).

Convert the angle to radians:

\(\theta=29.5413\ldots^\circ=0.515594\ldots\text{ radians}\).

Area of sector \(ABD\):

\(\text{Area}=\dfrac{1}{2}r^2\theta\).

\(\text{Area of sector }ABD=\dfrac{1}{2}(10^2)(0.515594\ldots)\).

\(\text{Area of sector }ABD=25.7797\ldots\text{ cm}^2\).

Therefore,

\(\text{Shaded area}=36.9788\ldots-25.7797\ldots\).

\(\text{Shaded area}=11.1991\ldots\text{ cm}^2\).

✅ Answer: \(11.2\text{ cm}^2\)