Question

If m∠A = 72°, m∠B = 32°, and c = 8, what are the measures of the remaining sides and angle?

Answer 1

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°

Methods and calculations used to obtain the above responses

Question 1:

If m∠A = 72°

m∠B = 32°

c = 8

Therefore;

m∠C = 180° - 72° - 32° = 76°

$\displaystyle \frac{a}{sin(72^{\circ})} = \mathbf{ \frac{8}{sin(76^{\circ})} }$

$\displaystyle a = \mathbf{\frac{8}{sin(76^{\circ})} \times sin(72^{\circ})} \approx 7.84$

$\displaystyle b = \frac{8}{sin(76^{\circ})} \times sin(32^{\circ}) \approx 4.37$

The correct option is therefore;

• A. m∠C = 76°, a ≈ 7.84, b ≈ 4.37

Question 2

The given parameters are;

$\displaystyle B = \mathbf{ \frac{\pi}{6} }$

$\displaystyle \frac{\pi}{6} = 30^{\circ}$

a = 20

b = 10

Therefore,  by the law of cosine we have;

• b² = a² + c² - 2·a·c·cos(B)

Which gives;

10² = 20² + c² - 2×20×c×cos(30°)

100 = 400 + c² - 20·c·(√3)

c² - 20·√3 ·c + 300 = 0

$\displaystyle c = \frac{20 \cdot \sqrt{3} \pm\sqrt{1200-4 \times 1 \times 300} }{2 \times 1} = \mathbf{ 10 \cdot \sqrt{3} }$

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;

$\displaystyle \frac{24}{sin(31^{\circ})} =\mathbf{\frac{x}{sin(42^{\circ})} }$

Where:

x = The distance from weather station A from the storm

Which gives;

$\displaystyle x = \frac{24}{sin(31^{\circ})} \times sin(42^{\circ}) \approx \mathbf{31.2 \ miles}$

• The distance from weather station A from the storm is D. 31.2 miles

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;

$\displaystyle \frac{b}{sin(71^{\circ})}= \mathbf{ \frac{9}{sin(52^{\circ})} }$

Therefore;

$\displaystyle b = \frac{9}{sin(52^{\circ})} \times sin(71^{\circ}) \approx \mathbf{ 10.799}$

• The correct option is; A. b = 10.799

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;

$\displaystyle \frac{200}{sin(110^{\circ})} = \frac{140}{sine \ of \ the \ angle \ at \ the \ hole, \ \phi}$

Therefore;

$\displaystyle sin(\phi) = \mathbf{ \frac{140}{200} \times sin(110^{\circ})} = 0.7 \times sin(110^{\circ})$

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°

• The angle that the golfer has between the spectator and the hole is C. 28.9°

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