Given: ∆ABC, AC = 12, AB = BC = 10 Find: m∠A, m∠C, m∠B
1 answer:
Answer:
- m∠A ≈ 53.13°
- m∠B ≈ 73.74°
- m∠C ≈ 53.13°
Step-by-step explanation:
An altitude to AC bisects it and creates two congruent right triangles. This lets you find ∠A = ∠C = arccos(6/10) ≈ 53.13°.
Since the sum of angles of a triangle is 180°, ∠B is the supplement of twice this angle, so is about 73.74°.
m∠A = m∠C ≈ 53.13°
m∠B ≈ 73.74°
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The mnemonic SOH CAH TOA reminds you of the relation between the adjacent side, hypotenuse, and trig function of an angle:
Cos = Adjacent/Hypotenuse
If the altitude from B bisects AC at X, triangle AXB is a right triangle with side AX adjacent to the angle A, and side AB as the hypotenuse. AX is half of AC, so has length 12/2 = 6, telling you the cosine of angle A is AX/AB = 6/10.
A diagram does not have to be sophisticated to be useful.
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Answer:
3.09 miles
Step-by-step explanation:
Given one distance and two angles, we will need to use the Law of Sines. For this, we need to know the internal angles of the triangle formed by the various bearing lines.
The angle between the bearings of A and B from the transmitter will be the difference of the reverse of the given bearings.
A from T = 39.3° +180° = 219.3°
B from T = 313.9° -180° = 133.9°
Then the angle at T between receivers is ...
219.3° -133.9° = 85.4°
The angle between A and T as measured at B will be ...
313.9° -270° = 43.9°
These angles and length AB can be used with the Law of Sines to find AT:
AT/sin(B) = AB/sin(T)
AT = AB(sin(B)/sin(T)) = (4.44 mi)·sin(43.9°)/sin(85.4°)
AT ≈ 3.09 mi
The distance of the transmitter from A is about 3.09 miles.
