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Serggg [28]
3 years ago
10

Line BC= 2.7cm, Line Ac= 1.1cm, Angle C = 90 Solve the tirangle

Mathematics
1 answer:
liraira [26]3 years ago
3 0
<span>The correct solution to this mathematical problem is AB2=BC2+AC2 or AB is a square root from the addition of the exponential values from both BC and AC. The final result is AB= 2.91547594742. It is easy to get the result by calculating the square root from the addition of the exponentiations of the values from the two sides of the triangle that we already have. We already have the value from BC=2,7 and AC=1,1, the angle C is 90 degrees. The equation comes from the old Pitagoras theorem which says that in a right triangle or triangle in which one of the angles is 90 degrees the exponential value of the hypotenuse is equal to the addition of the exponential values from the other two sides of this triangle. As for the other angles of this triangle, angle A and angle B we can find their values using this equation sinA=BC/AB and sinB=AC/AB or angle which are also applicable for right triangles. The function sin is a proportion from the opposite side of a right angle in the triangle and the hypotenuse of the same triangle. According to these equations the value of angle A is 67 degrees and angle B is 23 degrees. To verify the result from this equation we can use the rule that says the result from the addition of the three angles from each triangle should always be 180 degrees. In this case angle A + angle B + angle C = 67+23+90 = 180, which means that we got the right answer.</span>
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Step-by-step explanation:

expecting both 2x-15 and 3x are angles in radiant, let's draw a rhombus ABCD

∠ABC = 2x-15

∠ BCD = 3x

∠ABC + < BCD= π ( 180° in radiant)

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3 years ago
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3 years ago
Whitch three capital letters have rotational symmetry but do not have line symmetry
deff fn [24]
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So:
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This means that if we rotate the capital letter 180 deg, either clockwise or counterclockwise, it will still look the same

2. Must not have line symmetry.
If an object has line symmetry, it means that if you draw a line down the middle (in any way), it will be symmetrical on both sides. We need capital letters that do not fit that condition. 

Now we look at all capital letters. 
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But, we have to make sure they do not have line symmetry. If we draw a line right down the middle of H, I, O and X (**note, the have multiple lines of symmetry), they are symmetrical on both sides of the line. 

Now we are left with N, S, and Z
4 0
3 years ago
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