The laws of cosine and sine and the given parameters can be used to
calculate the measure of angles and distances.
Correct responses:
Question 1: A. m∠C = 76°, a = 7.84, b = 4.37
Question 2: C. 1 triangle
Question 3: D. 31.2 miles
Question 4: A. b = 10.799
Question 5: C. 28.9°
Question 1:
If m∠A = 72°
m∠B = 32°
c = 8
Therefore;
m∠C = 180° - 72° - 32° = 76°
[tex]\displaystyle \frac{a}{sin(72^{\circ})} = \mathbf{ \frac{8}{sin(76^{\circ})} }[/tex]
[tex]\displaystyle a = \mathbf{\frac{8}{sin(76^{\circ})} \times sin(72^{\circ})} \approx 7.84[/tex]
[tex]\displaystyle b = \frac{8}{sin(76^{\circ})} \times sin(32^{\circ}) \approx 4.37[/tex]
The correct option is therefore;
Question 2
The given parameters are;
[tex]\displaystyle B = \mathbf{ \frac{\pi}{6} }[/tex]
[tex]\displaystyle \frac{\pi}{6} = 30^{\circ}[/tex]
a = 20
b = 10
Therefore, by the law of cosine we have;
Which gives;
10² = 20² + c² - 2×20×c×cos(30°)
100 = 400 + c² - 20·c·(√3)
c² - 20·√3 ·c + 300 = 0
[tex]\displaystyle c = \frac{20 \cdot \sqrt{3} \pm\sqrt{1200-4 \times 1 \times 300} }{2 \times 1} = \mathbf{ 10 \cdot \sqrt{3} }[/tex]
Therefore, given that c has only one value, the three sides of the triangle are known, and the number of triangles unique triangles by Side-Side-Side description is; C. 1 triangle
Question 3
The distance between the weather station = 24 miles
The bearing of the storm from weather station A = N17°W
The bearing of the storm from weather station B = N48°W
The angle formed at point C in ΔABC is therefore;
108° - 42° - 107° = 31°
By sine rule, we have;
[tex]\displaystyle \frac{24}{sin(31^{\circ})} =\mathbf{\frac{x}{sin(42^{\circ})} }[/tex]
Where:
x = The distance from weather station A from the storm
Which gives;
[tex]\displaystyle x = \frac{24}{sin(31^{\circ})} \times sin(42^{\circ}) \approx \mathbf{31.2 \ miles}[/tex]
Question 4
∠A = 52°
∠C = 57°
Side BC = 9
Therefore;
∠B = 180° - 52° - 57° = 71°
∠B = 71°
According to the law of sines, we have;
[tex]\displaystyle \frac{b}{sin(71^{\circ})}= \mathbf{ \frac{9}{sin(52^{\circ})} }[/tex]
Therefore;
[tex]\displaystyle b = \frac{9}{sin(52^{\circ})} \times sin(71^{\circ}) \approx \mathbf{ 10.799}[/tex]
Question 5
Given:
Label of the vertex of the triangle formed are; The golfer, spectator, hole
Distance from the golfer to the hole = 200 yards
Distance from the golfer to the spectator = 140 yards
Vertex angle at the spectator = 110°
By using the law of sines, we have;
[tex]\displaystyle \frac{200}{sin(110^{\circ})} = \frac{140}{sine \ of \ the \ angle \ at \ the \ hole, \ \phi} [/tex]
Therefore;
[tex]\displaystyle sin(\phi) = \mathbf{ \frac{140}{200} \times sin(110^{\circ})} = 0.7 \times sin(110^{\circ})[/tex]
The angle at the hole, ∅ = arcsin(0.7×sin(110°)) ≈ 41.13°
Therefore;
The angle that the golfer has = 180° - 110° - 41.1° = 28.9°
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