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**sine Trigonometry cosine adjacent Ө Tangent Right Angle**

Sample Questions & Solutions Tangent cosine sine adjacent Right Angle Ө

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Pythagoras Theorem Given a = 3.25m & b = 3.45m calculate the length of c a² + b² = c² 3.25² ² = c² = c² = c² c = √22.466 c = 4.74m Answer c a b

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Question Copy each of the triangles below and label each of the sides and the angle. (c) A (b) A A (a) (e) A (f) A (d) A

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Answer Copy each of the triangles below and label each of the sides and the angle. (c) A (b) A A (a) hyp adj opp hyp opp opp hyp adj (e) A adj adj (f) A adj (d) A opp hyp adj opp opp hyp hyp

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Question Use your calculator to answer to 4 decimal places (a) cos 25° (b) sin 50° (c) tan 34°

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Answer Use your calculator to answer to 4 decimal places (a) cos 25° (b) sin 50° (c) tan 34° (a) cos 25 = = (b) sin = = (c) tan = =

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**Question Use your calculator to find the angles to the nearest degree**

(a) sin A = 0.83 (b) cos B = ¾ (c) tan A = 5/7

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**Answer Use your calculator to find the angles to the nearest degree**

(a) sin A = 0.83 (b) cos B = ¾ (c) tan A = 5/7 (a) 2ndF, sin, .83 = = 56° (b) 2ndF, cos, a/b, 3 ↓ 4 = = 41° (c) 2ndF, tan, a/b, 5 ↓ 7 = = 36°

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**Solving Right-angle Triangles**

In the triangle below, find the length of b The side we must find is the adjacent (b) We have the hypotenuse c = 10m The ratio that uses adj. & hyp. is cos So cos 60° = = 0.5 = b = 10(0.5) b = 5m Answer adjacent hypotenuse b 10 b 10 c = 10m a 60° b

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**Solving Right-angle Triangles**

Find the angle A to the nearest degree We have the opposite & the adjacent so we use tan ratio Tan A = = Tan A = Use calculator 2ndF, tan, a/b, 4 ↓ 3 = A = A = 53° Answer Opposite adjacent 4 3 4 3 A° 3m adjacent 4m opposite

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**Solving Right-angle Triangles**

Find the side c (hyp.) to 2 decimal places We are given angle & opposite to find hyp. sin 40° = = x c = 15 c = c = Answer 15 c 15 c 15 0.6428 c 15m 40° b

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**Solving Right-angle Triangles**

Given a section through a roof with 2 unequal pitches, calculate (a) the rise of the roof (b) the length of rafter c (c) the span b 5.42m c 40° 30° b

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**Solving Right-angle Triangles**

This question may seem complicated but we can simplify it by using our knowledge of right-angle triangles Firstly, divide triangle into 2 right-angled triangles (using the rise line) 5.42m c rise 40° 30° b

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**Solving Right-angle Triangles**

On the larger triangle we have the angle 30° & the hypotenuse 5.42m so we can use ratio sin 30° to find the opposite (rise) Sin = (soh) opposite hypotenuse 5.42m c 40° 30° b

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**Solving Right-angle Triangles**

Sin 30°= 0.5 = 0.5 x 5.42 = rise rise = 2.71m Answer (a) opposite hypotenuse rise 5.42 5.42m c 40° 30° b

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**Solving Right-angle Triangles**

So on the smaller triangle we now have the opposite (2.71m) and the angle 40° So to find adjacent (part of span) we use Tan Tan 40° = 0.839 = 0.839 x adj. = 2.71 adj. = 2.71 ÷ 0.839 = 3.23m Opposite adjacent 2.71 adjacent 5.42m c 30° 40° b adj.

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**Solving Right-angle Triangles**

Using Pythagoras we can find c a² + b² = c² 2.71² ² = c² = c² c² = c = √17.777 c = 4.216m Answer (b) 5.42m c 2.71m 40° 30° b 3.23m

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**Solving Right-angle Triangles**

Using Pythagoras we can find the rest of the span i.e. the base of the larger triangle a² + b² = c² 2.71² + b² = 5.42² b² = b² = – 7.344 b² = b = √22.032 b = 4.694m 5.42m c 2.71m 40° 30° b 3.23m

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**Solving Right-angle Triangles**

So = span Span = 7.924m Answer (c) 5.42m c 2.71m 40° 30° b 3.23m

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Notes - Trigonometry *I can solve right triangles in real world situations using sine, cosine and tangent. *I can solve right triangles in real world situations.

Notes - Trigonometry *I can solve right triangles in real world situations using sine, cosine and tangent. *I can solve right triangles in real world situations.

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