Question

w8 version (november 2025) name: test #4 - trigonometry part 1 section tota d1 determine the values of the trigonometric ratios for angles less than 360°; prove simple trigonometric identities; and solve problems using the primary trigonometric ratios, the sine law, and the cosine law; /35 please draw the special triangles for reference here. this is not worth marks. 1. answer the following questions in the appropriate blanks. 9 a) in which quadrant is the terminal arm of -303° found? b) determine the related acute angle of 193°. c) determine the principal angle of 456°. d) determine the second positive co-terminal angle to -57°. e) in which quadrant does the terminal arm of θ lie if cotθ and sinθ are both negative? f) determine the sign of the exact value of sin(-125°). (write positive or negative) g) which of the following is equivalent to sin 30°? i) sin(-30°) ii) cos 300° iii) cos 390° iv) cos(-120°) h) state the number of triangles that could exist: <a = 145°, a = 25 cm, b = 19 cm i) state the number of triangles that could exist: <a = 43°, a = 16.8 cm, b = 25.00 cm

Explanation:

Step1: Determine quadrant for \(-303^{\circ}\)

Add \(360^{\circ}\) to \(-303^{\circ}\): \(-303^{\circ}+360^{\circ} = 57^{\circ}\). Angles between \(0^{\circ}\) and \(90^{\circ}\) are in the first quadrant.

Step2: Find related acute angle for \(193^{\circ}\)

Since \(180^{\circ}<193^{\circ}<270^{\circ}\), the related acute angle is \(193^{\circ}-180^{\circ}=13^{\circ}\).

Step3: Calculate principal angle for \(456^{\circ}\)

Subtract \(360^{\circ}\) from \(456^{\circ}\): \(456^{\circ}-360^{\circ}=96^{\circ}\).

Step4: Find second positive co - terminal angle for \(-57^{\circ}\)

First positive co - terminal: \(-57^{\circ}+360^{\circ}=303^{\circ}\). Second positive co - terminal: \(303^{\circ}+360^{\circ}=663^{\circ}\).

Step5: Determine quadrant for \(\theta\) with \(\cot\theta<0\) and \(\sin\theta<0\)

\(\cot\theta=\frac{\cos\theta}{\sin\theta}<0\) and \(\sin\theta<0\) implies \(\cos\theta>0\). Angles with \(\sin\theta<0\) and \(\cos\theta>0\) are in the fourth quadrant.

Step6: Find sign of \(\sin(-125^{\circ})\)

\(\sin(-125^{\circ})=-\sin(125^{\circ})\). Since \(90^{\circ}<125^{\circ}<180^{\circ}\), \(\sin(125^{\circ})>0\), so \(\sin(-125^{\circ})\) is negative.

Step7: Check equivalence to \(\sin30^{\circ}\)

• \(\sin(-30^{\circ})=-\sin30^{\circ}\)
• \(\cos300^{\circ}=\cos(360^{\circ}-60^{\circ})=\cos60^{\circ}=\sin30^{\circ}\)
• \(\cos390^{\circ}=\cos(360^{\circ}+30^{\circ})=\cos30^{\circ}\)
• \(\cos(-120^{\circ})=\cos120^{\circ}=-\cos60^{\circ}\)

Step8: Determine number of triangles for \(A = 145^{\circ},a = 25,b = 19\)

Since \(A>90^{\circ}\) and \(a>b\), one triangle exists.

Step9: Determine number of triangles for \(A = 43^{\circ},a = 16.8,b = 25\)

Calculate \(b\sin A=25\sin43^{\circ}\approx25\times0.682=17.05\). Since \(a = 16.8<17.05\), no triangle exists.

Answer:

a) First quadrant
b) \(13^{\circ}\)
c) \(96^{\circ}\)
d) \(663^{\circ}\)
e) Fourth quadrant
f) Negative
g) ii) \(\cos300^{\circ}\)
h) \(1\)
i) \(0\)