Geometry Assignment: Trigonometry, Area, and Triangle Problems

This document presents a comprehensive solution set for a geometry assignment. The assignment includes problems involving trigonometric functions (sine, cosine, tangent) to find side lengths and angles in right triangles. It covers the calculation of areas of triangles, utilizing formulas and given dimensions. Furthermore, the solutions demonstrate the application of geometric principles to solve for unknown lengths within triangles using proportions and algebraic manipulation. The problems require the application of the Pythagorean theorem and trigonometric identities to find unknown values. The assignment also includes problems requiring the calculation of areas and the application of trigonometric ratios to solve for unknown side lengths and angles within triangles. The solutions provided are step-by-step, demonstrating the application of relevant formulas and concepts.

Geometry
Student Name:
Instructor Name:
Course Number:
29 March 2019

Q1:
a) Find AB to the nearest integer
Answer
cos A= Adjacent
Hypotenuse
cos 24= AB
22 =0.91354545764
AB=0.91354545764∗22=20.09800006808
≈ 20
b) Find BC to the nearest integer
Answer
sin A= Opposite
Hypotenuse
sin 24= BC
22 =0.40673664307
BC=22∗0.40673664307=8.94820614754
¿ 9
c) Area of triangle ABCD
Answer
Area=l∗w
¿ 20∗9
¿ 180 square units
Q2:
a) Find the length AD
Answer
x
6 = 6
x+ 9
( x +9 ) x=6∗6
x2+9 x=36
x2+9 x−36=0
x2+12 x−3 x−36=0
x ( x +12 ) −3 ( x +12 ) =0
( x +12 ) ( x−3 )=0
x=−12∨x=3
Since we can’t have a negative length thus x = 3
Length AD=x=3
b) Find the area of triangle ABC
Answer

Area=1
2∗A∗B
A=6
B=3+ 3+ 9=15
Area=1
2∗A∗B
¿ 1
2∗6∗15
¿ 3∗15
¿ 45 square units
c) Find the measure of angle A to the nearest degree
Answer
tan A= Opposite
Adjacent
¿ 6
3
¿ 2
A=tan−1 ( 2 )=63.43494882
≈ 63
Q3:
a) To the nearest tenth of a meter
i) TX
Answer
cos A= Adjacent
Hypotenuse
cos 27= TX
21.2
TX=21.2∗cos 27
TX=21.2∗0.89100652418
¿ 18.889338312616
≈ 18.9
ii) RT
Answer
tan A= Opposite
Adjacent
tan 45= RT
18.9
RT =18.9∗1
¿ 18.9
iii) RX
Answer

sin A= Opposite
Hypotenuse
sin 45= 18.9
RX
RX∗sin 45=18.9
RX= 18.9
sin 45 = 18.9
0.70710678118 =26.72863632909899
≈ 26.7
b) Finding the area of triangle RTX
Answer
Area=1
2∗18.9∗18.9=178.605
≈ 179 m2
Q4:
a) Find AD to the nearest hundredth
Answer
CD= √36−x2
x
√36−x2 = √36−x2
x+7
x ( x +7 )= ( √36−x2 )2
=36−x2
x2+7 x =36−x2
2 x2 +7 x−36=0
Solving for x yields
x=2.839389937671455∨x =−6.339389937671455
But since we can’t have negative length we consider
x=2.839389937671455
AD=x=2.839389937671455
≈ 2.84
b) Using the results from part a, we find the length of the altitude CD to the nearest tenth
Answer
CD= √ 36−x2
¿ √ 36−2.842
¿ √ 36−8.0656
¿ √ 27.9344
¿ 5.2853
≈ 5.3
Q5:
Answer

sin A= Opposite
Hypotenuse
sin 37= x
21 =0.60181502315
x=21∗0.60181502315=12.63811548615
¿ 12.6
Q6:
Answer
sin A= Opposite
Hypotenuse
sin A= 8
12 =0.6666666666666667
A=sin−1 ( 0.66667 )=41.8103149
≈ 42
Q7:
For angle A
sin A= Opposite
Hypotenuse = 8 cm
17 cm= 8
17
cos A= Adjacent
Hypotenuse= 15 cm
17 cm =15
17
tan A= Opposite
Adjacent = 8 cm
15 cm= 8
15
Thus the correct answer is C: cos A= 15
17