Given:
Right triangle with one angle 45°
To find:
The value of q and r.
Solution:
Opposite to θ = 16
Adjacent to θ = r
Hypotenuse = q
Using trigonometric ratio formula:
![$\tan \theta =\frac{\text{Opposite side to } \theta}{\text{Adjacent to } \theta}](https://img.qammunity.org/2021/formulas/mathematics/high-school/pca3c4biicl6gjfpqhx3yscg7xs262c2sy.png)
![$\tan 45^\circ =(16)/(r)](https://img.qammunity.org/2021/formulas/mathematics/high-school/axy4pp11ptu2qqoady0kqtoggxr0ahvxv7.png)
The value of tan 45° = 1
![$1 =(16)/(r)](https://img.qammunity.org/2021/formulas/mathematics/high-school/in89v6awmy2l6i74mu6ut0r52w0qx1bc3v.png)
Do cross multiplication, we get
r = 16
Using trigonometric ratio formula:
![$\sin \theta =\frac{\text{Opposite side to } \theta}{\text{Hypotenuse}}](https://img.qammunity.org/2021/formulas/mathematics/high-school/sqhbde3tgjpuwdiymu0hrpjf9ek8133u7m.png)
![$\sin 45^\circ =(16)/(q)](https://img.qammunity.org/2021/formulas/mathematics/high-school/9qhugmb4j58mt72usz19mteh4pb2akt6tc.png)
The value of sin 45° =
.
![$(1)/(√(2) ) =(16)/(q)](https://img.qammunity.org/2021/formulas/mathematics/high-school/cd221vv66zve1m1ktblgyh0rr6l7ge5frj.png)
Do cross multiplication, we get
![q=16√(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/v693llwwrazxgg7oex25pcmsich932v68g.png)
The value of r is 16 and the value of q is
.