The value of CD is 10.7 cm , Option C is the right answer.
<h3>What is a Line Segment ?</h3>
A line segment is a line of fixed measure.
From the figure , as Triangle ABC is a right angled triangle
sin 30 degree = (Perpendicular / Hypotenuse)
∠ B = 30° , AB= 10 cm
sin 30 = AC/10
AC = 5 cm.
To determine CD ,
In Triangle ADC ,
tan 25 = Perpendicular / Base
tan 25 = AC / CD
tan 25 = 5 / CD
CD = 5 / 0.4663
CD = 10.7 cm
Therefore the value of CD is 10.7 cm , Option C is the right answer.
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Answer:1.4348e-5 for dimensions and area is 160,000 sq ft. but thats only the first tree one...only one i can do without my brain giving out..good luck
Step-by-step explanation:
Question :-
- Find the Area of Rectangle , where the Lenght is 15 cm and its Breadth is 7 cm .
Answer :-
- Area of Rectangle is 105 cm² .

Diagram :-


Solution :-
» As per the provided information in the given question, we have been given that the Length of Rectangle is 15 cm . It's Breadth is given as 7 cm . And, we have been asked to calculate the Area of Rectangle.
For calculating the Area of Rectangle , we will use the Formula :-
Therefore , by Substituting the given values in the above Formula :-



Hence :-
- Area of Rectangle = 105 cm² .

Additional Information :-
![\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\ \underline{ { \textbf {\textsf \red{ \dag \: \: More \: Formulas \: \: \dag}}}} \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Square} = Side \times Side} \\ \\ \\ \footnotesize\bigstar \: \bf{Area \: _{Rectangle} = Lenght \times Breadth} \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Triangle} = \frac{1}{2} \times Base \times Height } \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Parallelogram} = Base \times Height} \\ \\ \\ \footnotesize \bigstar \: \bf{Area \: _{Trapezium} = \frac{1}{2} \times [ \: A + B \: ] \times Height } \\ \\ \\ \footnotesize \bigstar \: \bf {Area \: _{Rhombus} = \frac{1}{2} \times Diagonal \: 1 \times Diagonal \: 2}\end{array}}\end{gathered}\end{gathered}](https://tex.z-dn.net/?f=%20%5Cbegin%7Bgathered%7D%5Cbegin%7Bgathered%7D%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%20%5C%5C%20%5Cunderline%7B%20%7B%20%5Ctextbf%20%7B%5Ctextsf%20%5Cred%7B%20%5Cdag%20%5C%3A%20%20%5C%3A%20More%20%5C%3A%20Formulas%20%5C%3A%20%20%5C%3A%20%20%5Cdag%7D%7D%7D%7D%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5Cfootnotesize%20%5Cbigstar%20%20%5C%3A%20%20%5Cbf%7BArea%20%5C%3A%20_%7BSquare%7D%20%3D%20Side%20%5Ctimes%20Side%7D%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%20%5Cfootnotesize%5Cbigstar%20%20%5C%3A%20%20%5Cbf%7BArea%20%5C%3A%20_%7BRectangle%7D%20%3D%20Lenght%20%5Ctimes%20Breadth%7D%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5Cfootnotesize%20%5Cbigstar%20%5C%3A%20%20%5Cbf%7BArea%20%5C%3A%20_%7BTriangle%7D%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20Base%20%5Ctimes%20Height%20%7D%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5Cfootnotesize%20%5Cbigstar%20%5C%3A%20%20%5Cbf%7BArea%20%5C%3A%20_%7BParallelogram%7D%20%3D%20Base%20%5Ctimes%20Height%7D%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5Cfootnotesize%20%5Cbigstar%20%5C%3A%20%20%5Cbf%7BArea%20%5C%3A%20_%7BTrapezium%7D%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5B%20%5C%3A%20A%20%2B%20B%20%5C%3A%20%5D%20%5Ctimes%20Height%20%7D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cfootnotesize%20%5Cbigstar%20%5C%3A%20%5Cbf%20%7BArea%20%5C%3A%20_%7BRhombus%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20Diagonal%20%5C%3A%201%20%5Ctimes%20Diagonal%20%5C%3A%202%7D%5Cend%7Barray%7D%7D%5Cend%7Bgathered%7D%5Cend%7Bgathered%7D%20)
These are linear equations that can be solved simultaneously;
5m+3n=41
3m-6n=9 (multiplying the first equation by 3 and the second by 5)
15m+9n=123
15m-30n= 45 (subtracting the two equations)
39n = 78
n = 2,
and for m, 3m = 9+6(2)
= 21
m = 7
Therefore, n=2 and m=7