Question

In $\triangle ABC$, if $AB = 6\sqrt{3}$ cm, $AC = 12$ cm, and $BC = 6$ cm, then $\angle B$ will be:

• A. $120^\circ$
• B. $90^\circ$
• C. $60^\circ$
• D. $15^\circ$

Hint:

Use the cosine rule when all three sides of a triangle are known. If $\cos \theta = 0$, then the angle is $90^\circ$.

Answer 1

The Correct Option is C

Step 1: Recall the cosine rule.
In any triangle $ABC$, the cosine rule states that: \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] where side $a = BC$, side $b = AC$, and side $c = AB$.

Step 2: Substitute the given values.
Given: \[ AB = 6\sqrt{3}\ \text{cm}, \quad AC = 12\ \text{cm}, \quad BC = 6\ \text{cm} \] So, \[ a = 6, \quad b = 12, \quad c = 6\sqrt{3} \]
Step 3: Apply the cosine rule.
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] \[ \cos B = \frac{6^2 + (6\sqrt{3})^2 - 12^2}{2 \times 6 \times 6\sqrt{3}} \]
Step 4: Simplify the expression.
\[ \cos B = \frac{36 + 108 - 144}{72\sqrt{3}} = \frac{0}{72\sqrt{3}} = 0 \]
Step 5: Find angle $B$.
If $\cos B = 0$, then \[ B = 90^\circ \]
Step 6: Verification.
Recheck: \[ 36 + 108 = 144 \Rightarrow \cos B = 0 \Rightarrow B = 90^\circ \] Thus, the value of $\angle B$ is $90^\circ$.
Final Answer
Final Answer: \[ \boxed{90^\circ} \]