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Norma-Jean [14]
2 years ago
5

Draw as many different triangles as possible that have two sides of length 4 cm and a 45° angle. Clearly mark the side lengths a

nd angles given.
(points per unique attached triangle illustration that is fitting this criteria).


Please help!

Mathematics
1 answer:
Basile [38]2 years ago
5 0

Student Verified! ✅

Answer:

The answer is 3!

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please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me
mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are  equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH=\sqrt{(5-0)^{2}+(1-1)^{2}}=5 and

GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5

Now, mid point of JH=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

=(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)

Mid point of GI=(\frac{5}{2},1)

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

(GI)^{2}=(IJ)^{2}+(JG)^{2}                             (1)

Now, JI=\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20} and GJ=\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}

Substituting these values in (1), we get

5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }

25=20+5

25=25

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5  and HJ=\sqrt{(0-5)^2+(1-1)^2}=5 are equal, thus, GHIJ is a rectangle.

6 0
2 years ago
If you know trig please help! Will give brainliest!
kaheart [24]

Answer:

sin^2α

Step-by-step explanation:

I will just  pose  x instead  of alpha  here to make things simpler

\tan ^2\left(x\right)\left(2\cos ^2\left(x\right)+\sin ^2\left(x\right)-1\right)\\\\

we know that sin^2x = 1 -  cos^2x so...

=\left(-1+1-\cos ^2\left(x\right)+2\cos ^2\left(x\right)\right)\tan ^2\left(x\right)

=\cos ^2\left(x\right)\tan ^2\left(x\right)

we can rewrite using trigonometric identities (tan = sin/cos)...

=\left(\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)^2\cos ^2\left(x\right)\\= \sin ^2\left(x\right)

6 0
2 years ago
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