Solving Right Triangles

9 - 1 Solving Right Triangles

Samples of solving right triangles 1) For the right triangle ABC shown, find the side lengths to three significant digits and angles to tenths place.

To find / B, subtract / A from 90. B = 90 - 25 = 65 / B = 65^o To find side b, use / B. We can use: tan / B = opp/adj = b/40 tan 65 = b/40 40 tan 65 = b 85.8 = b To find side c, use / A. We can use: sin / A = opp/hyp = 40/c sin 25 = 40/c c sin 25 = 40 c = 40/sin 25 c = 94.6

2) Find the missing parts of the triangle shown below:

To find / A we can use tan tan / A = opp/adj tan / A = 50/30 / A = tan^-1(50/30) / A = 59.0^o To find / B, subtract / A from 90 / B = 90 - 59 = 31^o / B = 31^o To find c, use the pythagorean theorem. ________ c = \/ 30² + 50² c = 58.3

3) From a point 100 meters from the base of a tower, the angle of elevation to its top is 42^o. Find its height to three significant digits.

To find h, use tangent tan 38 = opp/adj tan 38 = h/100 100 tan 38 = h 78.1 = h The tower is 78.1 meters high.

4) From the top of a lighthouse 54 meters above the sea, the angle of depression of a buoy is 17^o. Find the horizontal distance from the buoy. Round to three significant digits.

To find x, use cotangent. cot 17 = adj/opp cot 17 = x/54 54 cot 17 = x 177 = x 177 meters to the buoy. We could also have used the tangent by finding the other acute angle. 90 - 17 = 73 tan 73 = x/54 54 tan 73 = x 177 = x Your choice!!

5) From the top of a building that overlooks a lake, Jack watches a boat sailing directly toward the building. He notes that the angle of depression now to the boat is 40^o and later it changes to 27^o. His room is 120 feet above the lake. How far did the boat travel during that period?

We need to calculate two distances and then subtract these two. Both distances we calculate will give us the distance from the boat to the shore at different times. So, by subtracting we will get the distance the boat traveled. Calculating the distance to the shore at the first observation of 27^o: Use tangent: tan 63 = opp/adj tan 63 = x/120 120 tan 63 = x 235.5 = x Now calculate the distance at 40^o: Use tangent again tan 50 = opp/adj tan 50 = y/120 120 tan 50 = y 143.0 = y Now subtract y from x and we will have the distance the boat traveled. 235.5 - 143.0 = 92.5 The boat traveled 92.5 feet.

That should give you some examples of the trig of the right triangle. Next section deals with calculating the area of a triangle without knowing the height.