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Reil [10]
3 years ago
6

How do I work this out???

Mathematics
1 answer:
Finger [1]3 years ago
7 0
Okay so form a right angle triangle and the angle of that is 65 as it is right angle so 90 - 25= 65. label the sides like the pic below. 

we have the hypothenuse as well as the angle and we are trying to find the o(opposite side) 

o/h =sin
sin 65=23/h
sin 65 = 23/x
23/sin 65=x 
x=25.4
how high the kite is above ground is :
= 25.4 +1.2
 =26.6

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Abigail drew a scale drawing of the oldest house in her town. The actual house is 64 feet long. Abigail's scale drawing is 8 inc
cestrela7 [59]

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The scale he used in drawing the house is determined by choosing a perfect scale:

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What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?
Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)  

We know that, distance between two points P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right) is given by  

P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Now,  

\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}

8 0
2 years ago
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