Question

exercise 10c
give answers to 3 s.f. where appropriate.
1 find the lengths of the sides marked by letters.
(a)
(b)
(c)
(d)

Explanation:

Step1: Use sine function for (a)

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For part (a), \(\theta = 30^{\circ}\), hypotenuse \(=10\mathrm{cm}\), and the side \(a\) is opposite to the \(30^{\circ}\) angle. So, \(\sin30^{\circ}=\frac{a}{10}\). Since \(\sin30^{\circ}=\frac{1}{2}\), then \(a = 10\times\sin30^{\circ}=10\times\frac{1}{2}=5\mathrm{cm}\).

Step2: Use cosine function for (b)

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For part (b), \(\theta = 56^{\circ}\), hypotenuse \(=35\mathrm{cm}\), and the side \(e\) is adjacent to the \(56^{\circ}\) angle. So, \(\cos56^{\circ}=\frac{e}{35}\). Then \(e = 35\times\cos56^{\circ}\approx35\times0.5592\approx19.67\approx19.7\mathrm{cm}\).

Step3: Use sine function for (c)

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For part (c), \(\theta = 39^{\circ}\), hypotenuse \(=40\mathrm{cm}\), and the side \(d\) is opposite to the \(39^{\circ}\) angle. So, \(\sin39^{\circ}=\frac{d}{40}\). Then \(d = 40\times\sin39^{\circ}\approx40\times0.6293\approx25.2\mathrm{cm}\).

Step4: Use cosine function for (d)

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For part (d), \(\theta = 70^{\circ}\), hypotenuse \(=10\mathrm{cm}\), and the side \(p\) is adjacent to the \(70^{\circ}\) angle. So, \(\cos70^{\circ}=\frac{p}{10}\). Then \(p = 10\times\cos70^{\circ}\approx10\times0.3420\approx3.42\mathrm{cm}\).

Answer:

(a) \(5\mathrm{cm}\)
(b) \(19.7\mathrm{cm}\)
(c) \(25.2\mathrm{cm}\)
(d) \(3.42\mathrm{cm}\)