Download Area of an oblique triangle {Trigonometry} {given two sides and an

ACT Practice
Trigonometry
Area of an oblique triangle {Trigonometry}
{given two sides and an included angle}
Area =
𝟏
𝐛𝐜 𝐬𝐢𝐧 𝐀
𝟐
Area of ⊿𝐀𝐁𝐂
1
30 = sin 70° xy
2
30 =
1
.93969 xy
2
60 = .93969xy
xy = 63.85
=
𝟏
𝐚𝐛 𝐬𝐢𝐧 𝐂
𝟐
=
𝟏
𝐚𝐜 𝐬𝐢𝐧 𝐁
𝟐
Area of ⊿𝐏𝐐𝐑
1
A = sin 110° xy
2
A=
1
.93969 xy
2
Areas are proven to be equal
using the property stated above!!!
© Mr. Sims
ACT Practice
Trigonometry
c
a
b
c² = a² + b² − 2ab cos ∠C
{given law of cosines}
a² = b² + c² − 2bc cos ∠A {put in terms of side a}
a² = 12² + 18² − 2 12 18 cos 40° {substituted values in}
a = 12² + 18² − 2 12 18 cos 40° {square root of each side}
© Mr. Sims
ACT Practice
Intermediate Algebra
The nth term of an arithmetic sequence can be represented by:
an = a + (n – 1)∙d
an is the nth term
a is the first term of the sequence
n is the number of term
d is the common difference
8, ___ , ___ , ___ , 13
10th term
6th term
the 10th term exceeds the 6th term by 5
10th term
equals 6th term plus 5
a + (10 – 1)∙d = a + (6 – 1)∙d + 5 {the 10th term is 5 more than the 6th term}
9d = 5d + 5 {subtracted a from each side and worked inside parentheses}
4d = 5 {subtracted 5d from each side}
d = 1.25 {divided each side by 4}
the common difference is 1.25
Work backwards to find the first 4 terms:
8 – 1.25 = 6.75 5th term
6.75 – 1.25 = 5.5 {4th term}
st four terms together
Add
the
1
rd
5.5 – 1.25 = 4.25 {3 term}
= 5.5 + 4.25 + 3 + 1.75
4.25 – 1.25 = 3 {2nd term}
3 – 1.25 = 1.75 {1st term}
= 14.5
© Mr. Sims
ACT Practice
Elementary Algebra
Work backwards from the solution to create a quadratic equation:
x = -3 or x = -3 {the graph crosses the x-axis, twice, at -3}
x + 3 = 0 or x + 3 = 0 {added 3 to both sides of each equation}
(x + 3)(x + 3) = 0 {both can be multiplied to equal 0}
x2 + 6x + 9 = 0 {squared the binomial}
x2 + mx + n = 0
© Mr. Sims
ACT Practice
Intermediate Algebra
5 units from -3 would be -8 and also 2
Absolute value is the distance away from zero.
|x + 3| = 5 {means (x + 3) is 5 units from zero}
x + 3 = 5 or x + 3 = -5 {(x + 3) could be 5 or -5}
x=2
or x = -8 {subtracted 3 from both sides of each equation}
The solution set to |x + 3| = 5 is
the set of real numbers that are 5 units from -3
© Mr. Sims
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