we know that
Perimeter of a triangle is equal to
P=a+b+c
where
a,b and c are the length sides of the triangle
<u>Find the perimeter of the triangles</u>
<u>Triangle N 1</u>
P=(x-2)+(x)+(3x+1)-------> P=5x-1
<u>Triangle N 2</u>
P=(2x-5)+(x+4)+(6x-7)------> P=9x-8
equate the perimeters
5x-1=9x-8--------> 9x-5x=-1+8-------> 4x=7
x=7/4------> x=1.75
therefore
the answer is
x=1.75
It is thousandths because it t is 3 plac s the the left
A)
To be similar triangles have to have equal angles
triangle ZDB'
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees
angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90
triangle ZDB'
DZB'+DB'Z=180-90=90
DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2
c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem
Answer:
<h2>
sin(01) = -32.70/13</h2>
Step-by-step explanation:
Given cos(01) = -30/13 where angle 01 is located in quadrant III. In the third quadrant, both cos and sin are negative. only tan is positive.
To calculate sin(01), we will apply the trigonometry expression as shown below;
cos(01) = -30/13
According to SOH CAH TOA;
cos(01) = adjacent/hypotenuse = -30/13
sin(01) = opposite/ hypotenuse
To get the hypotenuse, we will use the pythagoras theorem
sin(01) = 32.70/13
Since sin is also negaitve in the third quadrant, the value of
sin(01) = -32.70/13
