Question

a 12 - m support post supports a telephone pole that is leaning slightly towards it. the support post makes an angle of 50° with the ground and meets the telephone pole at a point 9.5 m from its base. what is the measure that the telephone pole makes with the ground?
65°
95°
75°
79°
question 6 (1 point)
a triangle has side lengths of 13 cm, 17 cm, and 19 cm. what is the measure of the smallest angle, to the nearest degree?
41°
42°
61°
77°

Explanation:

Step1: Identify the problem type

These are geometry - triangle problems. We will use the Law of Sines and Law of Cosines.

Step2: Solve the first problem

Let the triangle formed by the support post, the ground and the telephone - pole segment. Let the angle that the telephone pole makes with the ground be $\theta$.
We know one side $a = 9.5$ m, another side $b = 12$ m and the included - angle $\alpha=50^{\circ}$.
First, use the Law of Sines: $\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}$, where $\beta$ is the angle opposite to side $b$ in the triangle.
$\sin\beta=\frac{b\sin\alpha}{a}=\frac{12\sin50^{\circ}}{9.5}\approx\frac{12\times0.766}{9.5}=\frac{9.192}{9.5}\approx0.9676$.
$\beta=\sin^{- 1}(0.9676)\approx75^{\circ}$.
The angle that the telephone pole makes with the ground $\theta = 180^{\circ}-(90^{\circ}+(75^{\circ}-50^{\circ})) = 65^{\circ}$.

Step3: Solve the second problem

In a triangle with side lengths $a = 13$ cm, $b = 17$ cm, and $c = 19$ cm, the smallest angle is opposite the smallest side. Let the smallest side $a = 13$ cm.
Use the Law of Cosines: $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$.
$\cos A=\frac{17^{2}+19^{2}-13^{2}}{2\times17\times19}=\frac{289 + 361-169}{646}=\frac{481}{646}\approx0.7446$.
$A=\cos^{-1}(0.7446)\approx42^{\circ}$.

Answer:

1. A. $65^{\circ}$
2. B. $42^{\circ}$