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shutvik [7]
3 years ago
5

Find the value of x in the triangle shown below.

Mathematics
2 answers:
Ainat [17]3 years ago
7 0

Answer:

62

Step-by-step explanation:

Again, sorry for responding late.  I dont get notifications for comments.

Since this is an isosceles triangle.  So x, and the other angle are both base angles.  This means x and the other unnamed angle are the same, and 56 is the vertex angle.  Look at the attached photo.

Since all angles in a triangle add up to 180, we can solve for x.

2x + 56 = 180

x = 62

11Alexandr11 [23.1K]3 years ago
5 0

Answer:

60.77054139≈x. I don't know what decimal place you need to round it to though

Step-by-step explanation:

We can apply the Law of Sines to this situation.

The Law of Sine states that sinA/a=sinB/b=sinC/c

Thus, we can get the equation

sin(56)/5.7=sin(x)/6

Multiplying both sides by 6, we get

6sin(56)/5.7=sin(x)

Simplifying the left side, we get

0.872671129...=sin(x)

Lastly, multiplying the inverse of sine to both sides, we get

sin^-1(0.872671129...)=sin^-1*sin(x)

60.77054139≈x

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The vertices of ∆ABC are A(7,3),B(−8,6),C(0,−5). If ∆ABC is translated along the vector <−9,5>, what are the coordinates o
nexus9112 [7]

Answer:

The coordinates of the vertices of \bigtriangleup A'B'C' are A'=(-2,8), B'=(-17,-4) and C'=(-9,0).

Step-by-step explanation:

Let be A =(7,3), B = (-8,6) and C = (0,-5) the vertices of the original triangle, as well as \vec t = \left, the coordinates of the vertices of the new triangle are, respectively:

A' = A + \vec t

A' = (7,3)+(-9,5)

A' = (7-9,3+5)

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B' = B + \vec t

B' = (-8,6)+(-9,5)

B' = (-8-9,-9+5)

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C' = C + \vec t

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C'=(-9,0)

The coordinates of the vertices of \bigtriangleup A'B'C' are A'=(-2,8), B'=(-17,-4) and C'=(-9,0).

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