find x from the picture in attachment
I do not have access to graphics to create a an image, but
Label the points around the triangle, counter clockwise from the left most point. You should have A,B,C,D,E.
A being on the extreme left and E on the extreme right, with C being at the top. Point B is between A & C. Point D is between C & E.
Further clarifications:
Distances:
A-to-B is 16
B-to-C is 8
C-to-D is 4
D-to-E is 12
B-to-D is x+5
A-to-E is 3x+5
Cos(C) will refer to the angle $\displaystyle \angle ACE $ & $\displaystyle \angle BCD $
Cosine Law:
$\displaystyle c^2 = a^2 + b^2 - 2ab \cos\left(C\right) $
Equation1:
$\displaystyle \left( x + 5 \right)^2 = 8^2 + 4^2 - 2\times 8 \times4\times cos\left( C \right) $
Equation2:
$\displaystyle \left( 3 x +5 \right)^2 = \left( 16+8 \right)^2 + \left( 4+12 \right) ^2 - 2 \left( 16+8 \right) \left( 4+12 \right) cos \left( C \right)$
Two equations, unknown x and unknown angle C.
You can expand the LHS of equation1 $\displaystyle \left( x + 5 \right)^2 \; = \; x^2 + 10x + 25 $
and the same for equation2.
or re-write:
$\displaystyle \left( \; \left( x + 5 \right)^2 - 8^2 - 4^2 \right) \; \div \left( 2\times 8 \times4 \; \right) = cos\left( C \right) $
Either way will work.
My preference is to re-write the equations, and isolate the trig function (the cosine). This effectively gives you one equation with one unknown, which reduces to a simple quadratic equation.
Hope that helps.
Hello, beq!x!
That picture is way out-of-scale . . . and mine is no better.
There must be typo . . . The problem has no solution.
Code:A * * * 4 * * 8 * o E * o * * o * * o x + 5 * D o * 12 * * 16 * * * * * * * * * * * * * * * * * * * * * * B 3x + 5 C
I agree with Aidan . . . the Law of Cosines: .$\displaystyle \cos A \:=\:\frac{b^2+c^2-a^2}{2bc}$
In $\displaystyle \Delta ABC\!:\;\;\cos A \;=\;\frac{24^2 + 16^2 - (3x+5)^2}{2(24)(16)} \;\;{\color{blue}[1]}$
In $\displaystyle \Delta ADE\!:\;\;\cos A \;=\;\frac{8^2+4^2-(x+5)^2}{2(8)(4)} \;\;{\color{blue}[2]}$
Equate [1] and [2]: .$\displaystyle \frac{832 - (3x+5)^2}{768} \;=\;\frac{80 - (x+5)^2}{64}$
This simplifies to: .$\displaystyle x^2 + 30x + 49 \:=\:0$
Quadratic Formula: .$\displaystyle x \;=\;\frac{-30 \pm\sqrt{704}}{2} \;=\;\frac{-30\pm8\sqrt{11}}{2} \;=\;-15 \pm 4\sqrt{11}$
We have two roots: .$\displaystyle x \;\approx\;-1.7335,\;-28.2665$
But both roots yield a negative value for side $\displaystyle BC \,=\,3x+5$
Math ASimilar Triangles
The cat on the right is an enlargement of the cat on the left. They are exactly the same shape, but they are NOT the same size.These cats are similar figures.
Objects, such as these two cats, that have the same shape, but do not have the same size, are said to be "similar".
The mathematical symbol used to denote
similar is .Do you remember this symbol as "part" of the symbol for congruent??
Similar
Symbol
Definition: In mathematics, polygons are similar if their corresponding (matching) angles are equal and the ratio of their corresponding sides are in proportion.
(This definition allows for congruent figures to also be "similar",
where the ratio of the corresponding sides is 1:1.)
Facts about similar triangles:
is called the ratio of similitude
AATo show triangles are similar, it is sufficient to show that two angles of one triangle are congruent (equal) to two angles of the other triangle.
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