Given:
There are given the right angle triangle, ABC.
Where,
[tex]\begin{gathered} AB=5 \\ BC=9 \\ AC=? \end{gathered}[/tex]Explanation:
To find the missing side in the given right angle triangle, we need to use the Pythagoras theorem:
So,
From the Pythagoras theorem in triangle ABC:
[tex]AB^{^2}+BC^2=AC^2[/tex]Then,
Put the all values into the above formula:
[tex]\begin{gathered} AB^2+BC^2=AC^2 \\ 5^2+9^2=AC^2 \\ 25+81=AC^2 \\ 106=AC^2 \end{gathered}[/tex]Then,
[tex]\begin{gathered} AC^2=106 \\ AC=\sqrt{106} \\ AC=10.3 \end{gathered}[/tex]Therefore, the side AC is 10.3:
Now,
We need to find the missing angle in the given triangle ABC:
So,
To find the angle A, we need to use the formula of tan function:
So,
[tex]\begin{gathered} tanA=\frac{BC}{AB} \\ tanA=\frac{9}{5} \\ tanA=1.8 \end{gathered}[/tex]Then,
[tex]\begin{gathered} tanA=1.8 \\ A=tan^{-1}(1.8) \\ A=60.9 \end{gathered}[/tex]Now,
We need to find the value of angle C:
So,
To find the angle C, we need to use the angle of triangle rule:
So,
In a right-angled triangle, the addition of the two angles is equal to the 90 degrees.
Then,
[tex]\angle A+\angle C=90^{\circ}[/tex]Then,
Put the value of angle A into the above formula;
So,
[tex]\begin{gathered} \angle A+\angle C=90^{\circ} \\ 61^{\circ}+\angle C=90^{\circ} \\ \angle C=90^{\circ}-61^{\circ} \\ \angle C=29^{\circ} \end{gathered}[/tex]Final answer:
Hence, the missing side and the missing angles are shown below:
[tex]\begin{gathered} AC=10.3 \\ \angle A=61^{\circ} \\ \angle C=29^{\circ} \end{gathered}[/tex]
