Answer:
a. A = 47.3°, B = 42.7°, c = 70.8 units
b. x ≈ 17.3 units, Y = 60°, z ≈ 34.6 units
Explanation:
You want to solve the right triangles ...
a) ABC, where a = 52, b = 48, C = 90°
b) XYZ, where y = 30, X = 30°, Z = 90°
Right triangles
The relations you use to solve right triangles are ...
- the Pythagorean theorem: c² = a² +b²
- trig definitions, abbreviated SOH CAH TOA
- sum of angles is 180° (acute angles are complementary)
a. ∆ABC
The hypotenuse is given by ...
c² = a² +b²
c² = 52² +48² = 2704 +2304 = 5008
c = √5008 ≈ 70.767
Angle A is given by ...
Tan = Opposite/Adjacent . . . . . this is the TOA part of SOH CAH TOA
tan(A) = BC/AC = 52/48
A = arctan(52/48) ≈ 47.3°
B = 90° -47.3° = 42.7° . . . . . . . . . . acute angles are complementary
The solution is A = 47.3°, B = 42.7°, c = 70.8 units.
b. ∆XYZ
The missing angle is ...
Y = 90° -30° = 60°
The given side XZ is adjacent to the given angle X, so we can use the cosine function to find the hypotenuse XY.
Cos = Adjacent/Hypotenuse . . . . this is the CAH part of SOH CAH TOA
cos(30°) = 30/XY
XY = 30/cos(30°) ≈ 34.641
The remaining side YZ can be found several ways. We could use another trig relation, or we could use the Pythagorean theorem. Another trig relation requires less work with the calculator.
Sin = Opposite/Hypotenuse . . . . . the SOH part of SOH CAH TOA
sin(30°) = YZ/XY
YZ = XY·sin(30°) = 34.641·(1/2) ≈ 17.321
The solution is x ≈ 17.3, Y = 60°, z ≈ 34.6.
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Additional comments
In triangle XYZ, the sides opposite the angles are x, y, z. That is x = YZ, y = XZ, and z = XY. The problem statement also says YZ = h. Perhaps this is a misunderstanding, as the hypotenuse of this triangle is opposite the 90° angle at Z, so will be XY.
Triangle XYZ is a 30°-60°-90° triangle. This is one of two "special" right triangles with side lengths in ratios that are not difficult to remember. The ratios of the side lengths in this triangle are 1 : √3 : 2. The given side is the longer leg, so corresponds to √3. That means the short side (x=YZ) is 30/√3 = 10√3 ≈ 17.3, and the hypotenuse is double that.
(The other "special" right triangle is the isosceles 45°-45°-90° right triangle with sides in the ratios 1 : 1 : √2.) You will see these often.
There are a couple of other relations that are added to the list when you are solving triangles without a right angle.
The first two attachments show the result of using a triangle solver web application. The last attachment shows the calculator screen that has the computations we used. (Be sure the angle mode is degrees.)
We have rounded our results to tenths, for no particular reason. You may need to round differently for your assignment.
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