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Trigonometry Sine Rule Mr Porter A B C a c b PQ R p q r

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Definition: The Sine Rule In any triangle ABC ‘The ratio of each side to the sine of the opposite angle is CONSTANT. A B C a c b PQ R p q r For triangle ABCFor triangle PQR

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Example 1: Use the sine rule to find the value of x correct to 2 decimal places. C AB 8.3 cm x 81°43° When to apply the sine rule: Is there 2 sides and 2 angles (or more)? YES, then use the Sine Rule. Write down the sine rule for this triangle Label the triangle. a b c We do not need the ‘C’ ratio. Substitute A, a, B and b. Rearrange to make x the subject. Use calculator. Example 2: Find the size of α in ∆PQR in degrees and minutes. Q P R 22.4 cm 14 cm 118°27’ α When to apply the sine rule: Is there 2 sides and 2 angles (or more)? YES, then use the Sine Rule. Write down the sine rule for this triangle Label the triangle. p q r We do not need the ‘Q’ ratio.Substitute P, p, R and r. Rearrange to make sinα the subject. Evaluate RHS To FIND angle, use sin -1 (..) Convert to deg. & min.

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Ambiguous Case – Angles (Only) Example 1 : Use the sine rule to find the size of angle θ. C BA 41° 14.5 cm 9.8 cm θ When to apply the sine rule: Is there 2 sides and 2 angles (or more)? YES, then use the Sine Rule. Write down the sine rule for this triangle to find an angle Label the triangle. a b c We do not need the ‘C’ ratio.Substitute A, a, B and b. Rearrange to make sin θ the subject. Evaluate RHS To FIND angle, use sin -1 (..) But, could the triangle be drawn a different way? The answer is YES! C B A 41° 14.5 cm 9.8 cm θ A θ By supplementary angles: Which is correct, test the angle sum to 180°, to find the third angle, α. Case 1: Case 2: Hence, both answers are correct!

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Ambiguous Case – Angles (Only) Example 1 : Use the sine rule to find the size of angle θ. Q P R 122° θ 17 cm 8.5 cm When to apply the sine rule: Is there 2 sides and 2 angles (or more)? YES, then use the Sine Rule. Write down the sine rule for this triangle Label the triangle. p q r We do not need the ‘P’ ratio.Substitute Q, R, q and r. Rearrange to make sin θ the subject. Evaluate RHS To FIND angle, use sin -1 (..) But, could the triangle be drawn a different way? The answer is NO! Lets check the supplementary angle METHOD. Which is correct, test the angle sum to 180°, to find the third angle, α. Case 1: Case 2: Hence, the ONLY answer is correct!

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Example 3: Points L and H are two lighthouses 4 km apart on a dangerous rocky shore. The shoreline (LH) runs east–west. From a ship (B) at sea, the bearing of H is 320° and the bearing of L is 030°. a) Find the distance from the ship (B) to the lighthouse (L), to the nearest metre. b) What is the bearing of the ship (B) from the lighthouse at H? N 0° N 0° N 0° HL4 km B (ship) 320° 30° Use basic alternate angles in parallel line, Bearing and angle sum of a triangle to find all angles with the ∆BHL. 30° 60° 50° 40° 70° Re-draw diagram for clarity d H B L 4000 m 60°50° 70° When to apply the sine rule: Is there 2 sides and 2 angles (or more)? YES, then use the Sine Rule. Label the triangle. l b h We do not need the ‘L’ ratio.Substitute B, H, b and d, (h). Rearrange to make d the subject. Use calculator. Write down the (side) sine rule for this triangle (a) (b) From the original diagram : Bearing of Ship from Lighthouse H: H = 90°+50° H = 140°

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Example 4: To measure the height of a hill a surveyor took two angle of elevation measurements from points X and Y, 200 m apart in a straight line. The angle of elevation of the top of the hill from X was 5° and from Y was 8°. What is the height of the hill, correct to the nearest metre? Hence, the hill is 46 m high (nearest metre). h XY B T 200 m 5°8° Use basic angle sum of a triangle, exterior angle of a triangle and supplementary angles to find all angles with the ∆AYT And ∆YTB. 82° 172° 3° When to apply the sine rule: Is there 2 sides and 2 angles (or more)? YES, then use the Sine Rule to find TY= x. To find ‘h’, we need either length BY or TY! x Write down the (side) sine rule for this triangle We do not need the ‘Y’ ratio.Substitute X, T, t and x. Rearrange to make x the subject. Use calculator Do NOT ROUND OFF! To find h, use the right angle triangle ratio’s i.e. sin θ.

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Example 5: A wooden stake, S, is 13 m from a point, A, on a straight fence. SA makes an angle of 20° with the fence. If a hores is tethered to S by a 10 m rope, where, on the fence, is the nearest point to A at which it can graze? Fence Stake A BC S 20° 10 m 13 m 1) The closest point to A along the fence, is point B. Hence, we need to find distance AB. 2) Look at ∆ABS, to use the sine rule, need to find angle ABS or angle ASB. Write down the (angle) sine rule for this triangle Evaluate RHS To FIND angle, use sin -1 (..) From the diagram, it is obvious that angle B is Obtuse. B = 180 – 26° 24’ B = 153° 36’ We do not need the ‘S’ ratio.Substitute A, B, a and b. Rearrange to make sin B the subject. Then angle ASB = 182 – (153°36’ + 20) = 6° 24’ Now, apply the sine rule to find the length of AB. Write down the (side) sine rule for this triangle AB = m

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Example 6: Q, A and B (in that order) are in a straight line. The bearings of A and B from Q is 020°T. From a point P, 4 km from Q in a direction NW, the bearing of A and B are 112°T and 064°T respectively. Calculate the distance from A to B. Use basic alternate angles in parallel line, Bearing and angle sum of a triangle to find all angles in the diagram. 23 ° 45 ° 48 ° 92 ° 44 ° 88 ° d 112° N 0° P N 0° A 4 km N 0° Q N 0° B 45°20° 64° Not to scale! x y To find ‘d’, we need to work backward using the sine rule, meaning that we must find either x or y first. [There are several different solution!] In ∆POA, write down the (side) sine rule for this triangle We do not need the ‘P’ ratio.Substitute Q, A, q and y. Rearrange to make y the subject. Use calculator Do NOT ROUND OFF! In ∆PAB, write down the (side) sine rule. For this triangle. We do not need the ‘A’ ratio.Substitute P, B, y and d. Rearrange to make d the subject. Use calculator and y = Do NOT ROUND OFF!

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