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Using SohCahToa to solve right triangles:

Using SohCahToa to solve right triangles:. opposite. h. y. p. What does SohCahToa stand for?. o. t. e. n. u. s. x. e. opposite_. sin x =. a d j a c e n t. hypotenuse. Using reciprocal relationships, csc x = ? sec x = ? cot x = ?. adjacent_. cos x =. hypotenuse.

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Using SohCahToa to solve right triangles:

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  1. Using SohCahToa to solve right triangles: opposite h y p What does SohCahToa stand for? o t e n u s x e opposite_ sinx = a d j a c e n t hypotenuse Using reciprocal relationships, cscx = ? secx = ? cotx = ? adjacent_ cosx = hypotenuse opposite tanx = adjacent

  2. Examples: c a = 40 1. Find b to 3 significant digits: 28º b 2. The safety instructions for a 20ft ladder indicate that the ladder should not be inclined at more than a 70° angle with the ground. Suppose the ladder is leaned against a house at this angle. Find: a) the ladder’s height against the house, and b) how far the base of the ladder is from the house. 3. The highest tower in the world is in Toronto, Canada, and is 553m high. A 1.8m tall observer standing 100m from the base of the tower sights the top of the tower. The angle of elevation is the angle formed by the observer’s line of sight to the top of the tower and the horizontal. Find the measure of this angle to the nearest tenth of a degree. 4. A triangle has sides of lengths 8, 8, and 4. Find the measures of the angles of the triangle to the nearest tenth of a degree.

  3. Using SohCahToa, we can actually derive a formula to find the area of any triangle… if we have SAS. B We know that A = 1/2 bh a h C A b But also, from SohCahToa, sinC = h a Therefore, h = a sinC A = 1/2 ab sinC

  4. Examples: Two sides of a triangle have lengths 7cm and 4cm. The angle between the sides measures 73º. Find the area of the triangle. The area of triangle PQR is 15. If p = 5 and q = 10, find all possible measures of angle R. Adjacent sides of a parallelogram have lengths 6cm and 7cm, and the measure of the included angle is 30º. Find the area of the parallelogram. Suppose a triangle has two sides of length a and b. If the angle between these sides varies, what is the maximum possible area that the triangle can attain? What can you say about the minimum possible area?

  5. B The Law of Sines sinA sinB sinC a = = c In triangle ABC, a b c C A b Remember! When solving for an angle, there could be two possible answers because sinx = sin(180º - x). The Law of Cosines In triangle ABC (same as above): c2 = a2 + b2 - 2ab cosC a2 = b2 + c2 - 2bc cosA b2 = a2 + c2 - 2ac cosB

  6. Examples: A civil engineer wants to determine the distances from points A and B to an inaccessible point C. From direct measurement, the engineer knows that AB - 25m, the angle at A measures 110°, and the measure at angle B is 20°. Find AC and BC. Suppose that two sides of a triangle have lengths 3cm and 7cm and that the angle between them measures 130°. Find the length of the third side. The lengths of the sides of a triangle are 5, 10, and 12. Solve the triangle. In the diagram at the right, AB = 5, BD = 2, DC = 4, and CA = 7. Find AD. A C B D

  7. Once we have two angles in a triangle, the third angle can be found by subtracting the sum of the known two from 180°. But when do we use the law of sines and when do we use the law of cosines?

  8. Homework Problems this packet should be finishedPg. 334–335 #2, 7, 13, 15, 25Pg. 342–343 #2, 3, 7, 19, 20Pg. 347–348 #2–4, 17, 21, 22Pg. 352–353 #1, 2, 15, 16, 17

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