1.
Item 2. Solve 7sinθ+6=0 for 0≤θ<360.
Solution:7sinθ=-6 ⇒ sinθ=-6/7.Reference angle = sin⁻¹(6/7)≈59°.sin is negative in Quadrants III and IV.QIII: 180°+59°=239°.QIV: 360°−59°=301°.Answer: A
2.
Item 1. Find the value of θ if 0≤θ<360 and tanθ=-0.8391.
Solution:tanθ is negative, so θ is in Quadrants II and IV.Reference angle = tan⁻¹(0.8391) ≈ 40°.QII: 180°−40°=140°.QIV: 360°−40°=320°.Answer: B
3.
Item 3. Find the measure of ∠C.
Solution:Use the Sine Law (SSA case).sin51°/16.2 = sinB/14.5.sinB≈0.6956, so B≈44°.C=180°−51°−44°=85°.Answer: C
4.
Item 7. Determine the unknown side in the triangle.
Solution:Use Cosine Law.s²=20²+10²−2(20)(10)cos40°.s²≈193.58.s≈√193.58≈13.9 m.Answer: C
5.
Item 8. Charlie Brown sails on two bearings. How far is he from the starting point?
Solution:The included angle is 105°.Use Cosine Law:x²=6²+18²−2(6)(18)cos105°.x²≈415.9.x≈20.4 km.Answer: A
6.
Item 5. For △ABC, if ∠A=30° and c=20, determine how many triangles can be drawn if a=20.
Solution:Find the height:sin30°=h/20.h=20sin30°=10.Since side a equals the height (a=10 in the solution diagram), exactly one right triangle can be formed.Answer: B
7.
Item 4. Building angle of elevation problem. Determine the height of the building.
Solution:Angles in the triangle give 8° at the top.Use Sine Law:sin8°/105 = sin32°/x.x≈399.8 m.Then use the right triangle:sin40°=height/399.8.height≈257.0 m.Answer: D
8.
Item 6. For △DEF, if ∠D=30°, d=24 and f=48, determine how many triangles can be drawn.
Solution:Find the height:sin30°=h/48.h=24.Since d=24 equals the height, only one triangle is possible.The remaining angles are 90° and 60°.Answer: C
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