Question

1. calculate the length of side using the cosine law. show your work. 3 marks (triangle with sides 6.5, 8, angle 72° and side x) 2. calculate the measure of angle x using the cosine law. show your work. 3 marks (triangle with sides 7 cm, 9 cm, 10 cm and angle x) 3. solve the following triangle. show your work. 5 marks (triangle with angle a=52°, sides ab=18, ac=21, side a=bc)

Explanation:

Step1: Apply Cosine Law for side \(a\)

The Cosine Law formula for a side \(a\) in \(\triangle ABC\) (where \(\angle A = 52^\circ\), \(c = 18\), \(b = 21\)) is \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\). Substitute \(b = 21\), \(c = 18\), \(\angle A=52^\circ\):
\(a^{2}=21^{2}+18^{2}-2\times21\times18\times\cos(52^\circ)\)
Calculate \(21^{2}=441\), \(18^{2}=324\), \(\cos(52^\circ)\approx0.6157\):
\(a^{2}=441 + 324-2\times21\times18\times0.6157\)
\(a^{2}=765 - 756\times0.6157\)
\(a^{2}=765 - 465.47\)
\(a^{2}\approx299.53\)
\(a\approx\sqrt{299.53}\approx17.31\)

Step2: Apply Sine Law for \(\angle B\)

Sine Law: \(\frac{\sin(B)}{b}=\frac{\sin(A)}{a}\). So \(\sin(B)=\frac{b\sin(A)}{a}\). Substitute \(b = 21\), \(\sin(52^\circ)\approx0.7880\), \(a\approx17.31\):
\(\sin(B)=\frac{21\times0.7880}{17.31}\approx\frac{16.548}{17.31}\approx0.956\)
\(\angle B=\sin^{-1}(0.956)\approx73^\circ\) (or check with Cosine Law: \(\cos(B)=\frac{a^{2}+c^{2}-b^{2}}{2ac}\), substitute values to verify).

Step3: Find \(\angle C\)

Sum of angles in triangle: \(\angle C = 180^\circ-\angle A-\angle B\). Substitute \(\angle A = 52^\circ\), \(\angle B\approx73^\circ\):
\(\angle C=180 - 52 - 73 = 55^\circ\)

Answer:

\(\angle B\approx73^\circ\), \(\angle C = 55^\circ\), \(a\approx17.31\) (exact values may vary slightly with rounding)