From one ratio to all the ratios, and finding the sides of a right triangle
Mathematics Prep 3 Term 1, Unit 3, Lesson 4
GoalFind the other ratios from one given ratio, and find unknown sides of a right triangle from a ratio.
New words
given ratio · the one value of sin, cos or tan that we are toldscale factor k · the number that multiplies the simplest side lengthsPythagoras · a² + b² = c² in a right triangleexact value · a fraction, not a rounded decimal
Explanation
A ratio such as sin θ = 3/5 does not say the sides ARE 3 and 5; it says they are in the proportion 3 : 5. So call the opposite side 3k and the hypotenuse 5k. Pythagoras gives the third side: (5k)² − (3k)² = 16k², so the adjacent side is 4k. Now every ratio follows: cos θ = 4/5 and tan θ = 3/4. If one real length is given, it finds k. The method for any given ratio: (1) draw a right triangle and mark θ; (2) write the two sides given by the ratio as multiples of k; (3) find the third side with Pythagoras; (4) read the other ratios; (5) use a given length to find k and the real sides.
Worked examples
sin θ = 3/5 and the hypotenuse is 20 cm. Find the other sides, cos θ and tan θ. Opposite = 3k, hypotenuse = 5k = 20, so k = 4. Opposite = 12 cm, adjacent = 4k = 16 cm (check: 12² + 16² = 400 = 20²). cos θ = 16/20 = 4/5 and tan θ = 12/16 = 3/4.
cos θ = 5/13. Find sin θ and tan θ, and the hypotenuse if the adjacent side is 15 m. Adjacent 5k, hypotenuse 13k, so opposite = 12k (since 13² − 5² = 144). sin θ = 12/13 and tan θ = 12/5. If 5k = 15 then k = 3: hypotenuse 39 m and opposite 36 m. Check: 15² + 36² = 1521 = 39².
tan θ = 3/4 and the shorter leg is 9 cm. Find the perimeter. The legs are 3k = 9 and 4k, so k = 3: legs 9 and 12, hypotenuse 5k = 15. Perimeter = 9 + 12 + 15 = 36 cm.
Class activity
Ladder problem. A 10 m ladder leans on a wall and reaches 8 m up. Find sin θ (θ with the ground), then cos θ and how far the foot is from the wall. Check with Pythagoras.
Practice
1
sin θ = 5/13 and the hypotenuse is 26. Find the other sides, cos θ and tan θ.
k = 2: opposite 10, adjacent 24. cos θ = 12/13, tan θ = 5/12.
2
cos θ = 4/5 and the adjacent side is 20. Find the hypotenuse and the opposite side, and tan θ.
4k = 20, k = 5: hypotenuse 25, opposite 15, tan θ = 15/20 = 3/4.
3
tan θ = 5/12 and the adjacent side is 24. Find the other sides.
12k = 24, k = 2: opposite 10, hypotenuse 26.
4
sin θ = 8/17. Find cos θ and tan θ.
Adjacent = 15k (17² − 8² = 225). cos θ = 15/17, tan θ = 8/15.
5
A 10 m ladder reaches 8 m up a wall. How far is its foot from the wall?
sin θ = 8/10 = 4/5, so cos θ = 3/5 and the distance is 10 × 3/5 = 6 m (6² + 8² = 10²).
