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3 years ago

Does anybody know how to determine the angle again? of Q.B

Mathematics

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

Therefore the measur angle ACB

$m\angle ACB=61.93\°$

Step-by-step explanation:

Given:

ΔABC is a right Triangle at ∠A = 90°

AB = 15 ....Side Opposite to ∠ACB

AC = 8 ....Side Adjacent to ∠ACB

To Find:

m∠ACB = ?

Solution:

In Right Angle Triangle ABC, Tangent Identity we have

$\tan (\angle ACB) = \dfrac{\textrm{side opposite to angle ACB}}{\textrm{side adjacent to angle ACB}}$

Substituting the values we get

$\tan (\angle ACB) = \dfrac{AB}{AC}=\dfrac{15}{8}=1.875\\\\\angle ACB=\tan^{-1}(1.875)\\m\angle ACB=61.93\°$

Therefore the measur angle ACB

$m\angle ACB=61.93\°$

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3 years ago

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2 years ago

Read 2 more answers

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3 years ago

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.

8 0

3 years ago