The diagram shows triangle ABC - OCR - GCSE Maths - Question 20 - 2023 - Paper 4

To find angle BAC, we will use the Cosine Rule. The Cosine Rule states that:

$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$

Here:

• Let AB = c = 10.6 cm, AC = b = 12.5 cm, and BC = a = 8.2 cm.

We can rearrange the formula to find angle C:

$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$

Calculating each term:

• $a^2 = (8.2)^2 = 67.24$
• $b^2 = (12.5)^2 = 156.25$
• $c^2 = (10.6)^2 = 112.36$

Now substituting those values:

$\cos(C) = \frac{67.24 + 156.25 - 112.36}{2 \times 8.2 \times 12.5}$

Calculating the numerator: $67.24 + 156.25 - 112.36 = 111.13$

And the denominator: $2 \times 8.2 \times 12.5 = 205$

So, $\cos(C) = \frac{111.13}{205} \approx 0.541$

Now taking the arccos of this value to find C: $C = \cos^{-1}(0.541) \approx 57.8°$

Since angle BAC is opposite side BC, we apply the sines to find angle BAC. Using the Sine Rule:

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$ This gives us: $\frac{8.2}{\sin(A)} = \frac{10.6}{\sin(57.8°)}$

Calculating sin(57.8°): $\sin(57.8°) \approx 0.841$

Thus: $\frac{8.2}{\sin(A)} = \frac{10.6}{0.841}$

This leads to: $\sin(A) = \frac{8.2 \times 0.841}{10.6} \approx 0.634$

So, $A = \sin^{-1}(0.634) \approx 39.5°$

Rounding to one decimal place gives: $Angle \, BAC \approx 40.5°$