A boat is anchored 2.2 km due south of a lighthouse. Along the shore, the town of Beachton is 16.8 km due east of the lighthouse
. Enter a number in the box to correctly complete the statement. Round the answer to the nearest tenth. The distance from the boat to Beachton is 16 point 9 km
1 answer:
Answer:
16.9 km
Step-by-step explanation:
From the diagram attached, we can see that the problem forms a triangle.
So, to find the distance(d) from the boat to Beachton , we can make use of Pythagoreas theorem.
Thus;
d² = 16.8² + 2.2²
d² = 282.24 + 4.84
d² = 287.08
d = √287.08
d = 16.9434
Approximating to the nearest tenth, we have;
d ≈ 16.9 km
You might be interested in
Diagram for part A and part C is attached herewith.
Solving for part B:-
Given is the circle O. We start with drawing two radii OA and OC, then we join two points A and C to make a chord AC of the circle. Now The radius of the circle intersects the chord AC at point B such that it bisects AC into two equal parts AB and BC. Now we have two triangles ΔOBA and ΔOBC. In these two triangles, we have OA=OC (radii of circle), OB=OB (reflexive property), and BA=BC (given in the question). Using SSS congruency of triangles we can say ΔOBA≡ΔOBC and using CPCTC, we can conclude ∡OBA=∡OBC (=90°). Hence OB⊥AC i.e. OB is perpendicular to AC.
Solving for part D:-
Given is the circle O. We start with drawing two radii OA and OC, then we join two points A and C to make a chord AC of the circle. Now The radius of the circle intersects the chord AC at point B such that AB is perpendicular to AC i.e. ∡B=90°. Now we have two Right triangles ΔOBA and ΔOBC. In these two triangles, we have OA=OC (radii of circle), OB=OB (reflexive property). Using HL congruency of right triangles we can say ΔOBA≡ΔOBC and using CPCTC, we can conclude BA=BC. Hence OB bisects AC into AB=BC.
