Question

1. if \\( \tan b=\frac{11}{15} \\), find the measure of b. 1 mark 7. if \\( \cos a=-0.56 \\), find the measure of a. 1 mark 8. for each of the following triangles, find the measure indicated using primary trigonometric ratios, the sine law or the cosine law. a) find side a 3 marks b) find angle a 3 marks c) find angle \\( \theta \\) 3 marks

Explanation:

Step1: Find side \(a\) in part a)

Use the cosine ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\).
Here \(\theta = 28^{\circ}\), adjacent side is \(a\) and hypotenuse is \(8.0\) cm.
So, \(\cos28^{\circ}=\frac{a}{8.0}\).
Then \(a = 8.0\times\cos28^{\circ}\).
Using a calculator, \(\cos28^{\circ}\approx0.8829\).
So \(a\approx8.0\times0.8829 = 7.0632\approx7.1\) cm.

Step2: Find angle \(A\) in part b)

Use the sine law \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).
Here \(a = 11\) mm, \(b = 16\) mm, \(B = 96^{\circ}\).
So \(\frac{11}{\sin A}=\frac{16}{\sin96^{\circ}}\).
First, \(\sin96^{\circ}\approx0.9945\).
Then \(\sin A=\frac{11\times\sin96^{\circ}}{16}\).
\(\sin A=\frac{11\times0.9945}{16}\approx\frac{10.9395}{16}=0.6837\).
Using the inverse - sine function \(A=\sin^{-1}(0.6837)\approx43.1^{\circ}\).

Step3: Find angle \(\theta\) in part c)

Use the cosine law \(c^{2}=a^{2}+b^{2}-2ab\cos C\).
Let \(a = 15\) km, \(b = 15\) km, \(c = 10\) km.
\(10^{2}=15^{2}+15^{2}-2\times15\times15\times\cos\theta\).
\(100 = 225 + 225-450\cos\theta\).
\(100 = 450-450\cos\theta\).
\(450\cos\theta=450 - 100=350\).
\(\cos\theta=\frac{350}{450}=\frac{7}{9}\approx0.7778\).
Using the inverse - cosine function \(\theta=\cos^{-1}(0.7778)\approx38.9^{\circ}\).

Answer:

a) \(7.1\) cm
b) \(43.1^{\circ}\)
c) \(38.9^{\circ}\)