PLEASE HELP Solve the right triangle.
Round your answers to the nearest tenth.

Answer:
Step-by-step explanation:
Here we are given with a right angled traingle and we are interested in finding
Firstly we can use the ratio of sine to find the side a , as ;
[tex]\longrightarrow \sin\theta =\dfrac{p}{h} [/tex]
Substituting the respective values,
[tex]\longrightarrow \sin50^\circ =\dfrac{a}{14} [/tex]
Put on the value of sin50° = 0.766 ;
[tex]\longrightarrow 0.766 =\dfrac{a}{14} [/tex]
Cross multiply ,
[tex]\longrightarrow a = 14\times 0.766[/tex]
Simplify,
[tex]\longrightarrow \underline{\underline{\boldsymbol{ a = 10.724 \approx 10.8}}} [/tex]
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Now for finding b , we may use the ratio of cosine as ,
[tex]\longrightarrow \cos\theta =\dfrac{b}{h}[/tex]
Substituting the respective values,
[tex]\longrightarrow \cos50^\circ =\dfrac{b}{14}[/tex]
Put on the value of cos50° = 0.642 ;
[tex]\longrightarrow 0.642 =\dfrac{b}{14} [/tex]
Cross multiply,
[tex]\longrightarrow b =14\times 0.642[/tex]
Simplify,
[tex]\longrightarrow \underline{\underline{\boldsymbol{ b = 8.988 \approx 9}}}[/tex]
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Again we know that the sum of three angles of a triangle is 180° . So that;
[tex]\longrightarrow A +50^\circ +90^\circ =180^\circ [/tex]
Solve for A ,
[tex]\longrightarrow A =180^\circ -140^\circ[/tex]
[tex]\longrightarrow \underline{\underline{\boldsymbol{ A = 40^\circ }}}[/tex]
And we are done!