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ollegr [7]
3 years ago
7

Quadrilateral ABCD is inscribed in circle 0 What is m<A

Mathematics
1 answer:
valentinak56 [21]3 years ago
6 0
We know that
 A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. ( Inscribed Quadrilateral Theorem)
so
m∠B+m∠D=180°
2x+(3x-5)=180
5x=180+5
5x=185
x=185/5
x=37°

m∠A=x+5-----> 37+5------> 42°

the answer is
m∠A is 42°

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Solve the simultaneous equations<br><br> 3x + 2y = 30<br> 3x - y = 21
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Answer:

x = 8 , y = 3

Step-by-step explanation:

3x + 2y = 30

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3x + 2 (3x - 21) = 30

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3x + 6x = 30 + 42

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Step-by-step explanation:

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3 years ago
For questions 3 – 5, an airplane is heading south at an airspeed of 540 km/hr, but there is a wind blowing from the northeast at
GuDViN [60]

First, let's calculate the horizontal and vertical components of the wind speed (W) and the airplane speed (A), knowing that south is a bearing of 270° and northeast is a bearing of 45°:

\begin{gathered} W_x=W\cos45°\\ \\ W_x=50\cdot0.707\\ \\ W_x=35.35\\ \\ \\ \\ W_y=W\sin45°\\ \\ W_y=50\cdot0.707\\ \\ W_y=35.35 \end{gathered}\begin{gathered} A_x=A\cos270°\\ \\ A_x=540\cdot0\\ \\ A_x=0\\ \\ \\ \\ A_y=A\sin270°\\ \\ A_y=540\cdot(-1)\\ \\ A_y=-540 \end{gathered}

Now, let's add the components of the same direction:

\begin{gathered} V_x=W_x+A_x=35.35+0=35.35\\ \\ V_y=W_y+A_y=35.35-540=-504.65 \end{gathered}

To find the resultant bearing (theta), we can use the formula below:

\begin{gathered} \theta=\tan^{-1}(\frac{V_y}{V_x})\\ \\ \theta=\tan^{-1}(\frac{-504.65}{35.35})\\ \\ \theta=-86° \end{gathered}

The angle -86° is equivalent to -86 + 360 = 274°.

Therefore the correct option is b.

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