Take a bow

Geometry Level 5

You're lost amidst the jungle. Your bow has broken when you fell a ravine. The only thing you carry with you is a pocket knife, a tape-measure and some arrows in a bag. It's darkening and you can feel a chill in your spine, since fierce animals had started to roam by their habitat.

You climb a tree, find some strong bindweed and cogitate if can make a bow to your arrows.

Suppose you had found bindweed enough to weave a rope of 50 c m 50 cm . You're a skilled archer and know that for maximum precision of an arrow shot the arrow must have 2 3 \frac {2} {3} of the bow length and that distance between the middle point of the bow and the middle point of the rope must be 1 3 \frac {1} {3} of the arrow length. Your arrows have 25 c m 25 cm .

What must be the opening (in degrees) of the bow? Give your answer to 2 decimal places.

Details and assumptions

  • A bow is intended to be a circular section, which points have all the same distances from a single point, the centre of a circumference. Radius is the distance between this single point and any point of the bow;

  • If you need some trigonometrical functions, sin ⁡ ( 53.13 º ) = 0.8 \sin (53.13º)=0.8


The answer is 73.74.

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Mikael Marcondes by Mikael Marcondes , 24, Brazil

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1 solution

Mikael Marcondes
Jul 28, 2014

The distance between the centre of the circumference and the middle point of the rope can be expressed as r − f r-f , where f f is the distance between the middle of the rope and the middle of the bow. This distance is the side of a right triangle which other two sides are r r and c 2 \frac {c}{2} . Hence, replacing the measures given, comes:

r 2 − c 2 4 = ( r − f ) 2 r^{2} - \frac {c^{2}}{4} = (r-f)^{2}

r 2 − 5 0 2 4 = ( r − 25 3 ) 2 r^{2} - \frac {50^{2}}{4} = (r- \frac{25}{3})^{2}

r 2 − 625 = r 2 − 2 × r × 25 3 + 625 9 r^{2} - 625 = r^{2}- 2\times r \times \frac{25}{3} + \frac {625}{9}

r = 125 3 r = \frac {125}{3}

The cosine of the angle adjacent to the r − f r-f side is:

cos ⁡ θ = r − f r = 125 − 25 3 125 3 = 0.8 \cos \theta = \frac {r-f}{r} = \frac {\frac {125-25} {3}} {\frac {125} {3}} = 0.8

And the opening is:

2 × arccos ⁡ 0.8 = 2 × ( 90 º − arcsin ⁡ 0.8 ) = 2 × ( 90 º − 53.13 º ) = 73.74 º 2 \times \arccos {0.8} = 2 \times (90º-\arcsin {0.8})= 2 \times (90º-53.13º) = \boxed{73.74º}

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