Sagot :
[tex]\bold {SOLVING \: OBLIQUE \: TRIANGLE}[/tex]
[tex]\bold {Given:}[/tex]
- P = 78°
- R = 60°
- p = 16 in.
[tex]\bold {Unknown:}[/tex]
- angle Q
- side r
- side q
[tex]\bold {Solution:}[/tex]
Solving for ∠Q:
The sum of the interior angles of a triangle is 180°, thus,
[tex] \begin{aligned} \large \tt m \angle P + m \angle R + m \angle Q = 180° \\ \\ \large \tt 78° + 60° + m\angle Q = 180°\end{aligned} \\ \\ \large \tt m \angle Q=180°–78° - 60° \\ \\ \large \green{\boxed{\tt m \angle Q=42°}}[/tex]
[tex]\\[/tex]
Solving for r:
Using the Law of Sines,
[tex] \large \tt \frac{sinP}{p} = \frac{sinR}{r} \\ \\ \large \tt \frac{sin78}{16} = \frac{sin60}{r} \\ \\ \large \tt (sin78)(r) = (sin60)(16) \\ \\ \large \tt \frac{ \cancel{(sin78)}(r)}{ \cancel{(sin78)}} = \frac{(sin60)(16)}{(sin78)} \\ \\ \large\tt r=\frac{(sin60)(16)}{(sin78)} \\ \\ \large \green{ \boxed{\tt r =14.166in }}[/tex]
[tex]\\[/tex]
Solving for q:
Using the Law of Sines,
[tex] \large \tt \frac{sinP}{p} = \frac{sinQ}{q} \\ \\ \large \tt \frac{sin78}{16} = \frac{sin42}{q} \\ \\ \large \tt (sin78)(q) = (sin42)(16) \\ \\ \large \tt \frac{ \cancel{(sin78)}(q)}{ \cancel{(sin78)}} = \frac{(sin42)(16)}{(sin78)} \\ \\ \large\tt q=\frac{(sin42)(16)}{(sin78)} \\ \\ \large \green{ \boxed{\tt q =10.945in }}[/tex]
[tex]\\[/tex]
[tex]\bold {Final\:Answer:}[/tex] [tex]{\boxed{\begin{array}{l} \large \tt Q= \green{42°}\\ \\ \large \tt r=\green{14.166 \: in.} \\ \\ \large \tt q = \green{10.945 \: in.} \end{array}}}[/tex]
[tex]\\ \\[/tex]
#CarryOnLearning