Bearings & Triangles Quiz

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Cosines and sines with bearings

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JOCELYN AMOYO

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Froggy Jumps Bearings & Triangles QuizVersión en línea Cosines and sines with bearings por JOCELYN AMOYO 1 From point A, AB=25 km on bearing 040°. From A to C, AC=18 km on bearing 120°. Find BC (use the law of cosines). a BC ≈ 20.5 km b BC ≈ 36.5 km c BC ≈ 28.2 km 2 AB=12 km bearing 030°. AC=9 km bearing 090°. What is BC? (difference in bearing is 60°) a BC ≈ 15.4 km b BC ≈ 6.2 km c BC ≈ 10.8 km 3 AB=15 km bearing 045°. AC=20 km bearing 150°. Angle between them is 105°. Find BC. a BC ≈ 32.0 km b BC ≈ 27.9 km c BC ≈ 23.1 km 4 Two sides 7 km and 5 km with included bearing angle 40°. Use law of cosines to get BC. a BC ≈ 9.0 km b BC ≈ 11.7 km c BC ≈ 6.1 km 5 B and C are 13 km apart. From A, AB=9 km on bearing 010°, AC=12 km on bearing 260°. Find angle BAC using bearings then compute BC (cos rule). a BC ≈ 11.2 km b BC ≈ 17.0 km c BC ≈ 21.4 km 6 AB=11 km bearing 350°. AC=14 km bearing 100°. Distance BC by cosine rule? a BC ≈ 18.3 km b BC ≈ 12.7 km c BC ≈ 25.0 km 7 Using the sine rule, if AB=9 km opposite angle B and AC=12 km opposite angle C with angle A from bearings 60°, find BC. a BC ≈ 11.3 km b BC ≈ 14.9 km c BC ≈ 7.8 km 8 From A, AB=8 km bearing 015°, AC=6 km bearing 075°. Find BC using sine rule after getting angle A from bearing difference. a BC ≈ 9.5 km b BC ≈ 12.4 km c BC ≈ 5.6 km 9 AB=20 km bearing 120°, AC=15 km bearing 300°. Angle between paths is 180°−(120−300 mod 360). Compute BC. a BC ≈ 25.0 km b BC ≈ 18.1 km c BC ≈ 32.5 km 10 AB=14 km bearing 050°, AC=9 km bearing 170°. Solve BC by cosine rule. a BC ≈ 16.2 km b BC ≈ 8.3 km c BC ≈ 12.7 km

Froggy Jumps

Bearings & Triangles QuizVersión en línea

Cosines and sines with bearings

por JOCELYN AMOYO

From point A, AB=25 km on bearing 040°. From A to C, AC=18 km on bearing 120°. Find BC (use the law of cosines).

BC ≈ 20.5 km

BC ≈ 36.5 km

BC ≈ 28.2 km

AB=12 km bearing 030°. AC=9 km bearing 090°. What is BC? (difference in bearing is 60°)

BC ≈ 15.4 km

BC ≈ 6.2 km

BC ≈ 10.8 km

AB=15 km bearing 045°. AC=20 km bearing 150°. Angle between them is 105°. Find BC.

BC ≈ 32.0 km

BC ≈ 27.9 km

BC ≈ 23.1 km

Two sides 7 km and 5 km with included bearing angle 40°. Use law of cosines to get BC.

BC ≈ 9.0 km

BC ≈ 11.7 km

BC ≈ 6.1 km

B and C are 13 km apart. From A, AB=9 km on bearing 010°, AC=12 km on bearing 260°. Find angle BAC using bearings then compute BC (cos rule).

BC ≈ 11.2 km

BC ≈ 17.0 km

BC ≈ 21.4 km

AB=11 km bearing 350°. AC=14 km bearing 100°. Distance BC by cosine rule?

BC ≈ 18.3 km

BC ≈ 12.7 km

BC ≈ 25.0 km

Using the sine rule, if AB=9 km opposite angle B and AC=12 km opposite angle C with angle A from bearings 60°, find BC.

BC ≈ 11.3 km

BC ≈ 14.9 km

BC ≈ 7.8 km

From A, AB=8 km bearing 015°, AC=6 km bearing 075°. Find BC using sine rule after getting angle A from bearing difference.

BC ≈ 9.5 km

BC ≈ 12.4 km

BC ≈ 5.6 km

AB=20 km bearing 120°, AC=15 km bearing 300°. Angle between paths is 180°−(120−300 mod 360). Compute BC.

BC ≈ 25.0 km

BC ≈ 18.1 km

BC ≈ 32.5 km

AB=14 km bearing 050°, AC=9 km bearing 170°. Solve BC by cosine rule.

BC ≈ 16.2 km

BC ≈ 8.3 km

BC ≈ 12.7 km

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Bearings & Triangles Quiz

Froggy Jumps

Sobre esta actividad

Creada por

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Bearings & Triangles QuizVersión en línea

por JOCELYN AMOYO

From point A, AB=25 km on bearing 040°. From A to C, AC=18 km on bearing 120°. Find BC (use the law of cosines).

AB=12 km bearing 030°. AC=9 km bearing 090°. What is BC? (difference in bearing is 60°)

AB=15 km bearing 045°. AC=20 km bearing 150°. Angle between them is 105°. Find BC.

Two sides 7 km and 5 km with included bearing angle 40°. Use law of cosines to get BC.

B and C are 13 km apart. From A, AB=9 km on bearing 010°, AC=12 km on bearing 260°. Find angle BAC using bearings then compute BC (cos rule).

AB=11 km bearing 350°. AC=14 km bearing 100°. Distance BC by cosine rule?

Using the sine rule, if AB=9 km opposite angle B and AC=12 km opposite angle C with angle A from bearings 60°, find BC.

From A, AB=8 km bearing 015°, AC=6 km bearing 075°. Find BC using sine rule after getting angle A from bearing difference.

AB=20 km bearing 120°, AC=15 km bearing 300°. Angle between paths is 180°−(120−300 mod 360). Compute BC.

AB=14 km bearing 050°, AC=9 km bearing 170°. Solve BC by cosine rule.

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