IBDP Maths analysis and approaches Topic: SL 3.2 :Use of sine, cosine and tangent ratios HL Paper 1

Detail Solution

 Step 1: Use the formula for the area of a triangle based on two sides and the included angle.
The area \( A \) of triangle \( ABC \) can be calculated using the formula:
\[
Area = \frac{1}{2} \times AB \times AC \times \sin BAC
\]
We know \( AB = \sqrt{6} \) cm, \( AC = 10 \) cm, and we need to find \( \sin BAC \) using the cosine value.

Step 2: Calculate \( \sin BAC \) using the Pythagorean identity.
We know that:
\[
\sin^2 BAC + \cos^2 BAC = 1
\]
Substituting \( \cos BAC = \frac{1}{5} \):
\[
\sin^2 BAC + \left(\frac{1}{5}\right)^2 = 1
\]
\[
\sin^2 BAC + \frac{1}{25} = 1
\]
\[
\sin^2 BAC = 1 – \frac{1}{25} = \frac{25}{25} – \frac{1}{25} = \frac{24}{25}
\]
Taking the square root gives:
\[
\sin BAC = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5}
\]

Step 3: Substitute the values into the area formula.
Now substituting \( AB \), \( AC \), and \( \sin BAC \) into the area formula:
\[
Area = \frac{1}{2} \times \sqrt{6} \times 10 \times \frac{2\sqrt{6}}{5}
\]
Calculating this gives:
\[
Area= \frac{1}{2} \times 10 \times \frac{2 \times 6}{5}
\]
\[
Area  = \frac{1}{2} \times 10 \times \frac{12}{5} = \frac{10 \times 12}{10} = 12 \text{ cm}^2
\]

————Markscheme—————–

Using the formula for the area of a triangle:
\( \text{Area} = \frac{1}{2} \times AB \times AC \times \sin BAC \)
First, find \( \sin BAC \):
\( \sin BAC = \sqrt{1 – \left(\frac{1}{5}\right)^2} = \frac{\sqrt{24}}{5} \)
Then, \( \text{Area} = \frac{1}{2} \times \sqrt{6} \times 10 \times \frac{\sqrt{24}}{5} = 12 \) cm²