Statement In a triangle the sides b=9 cmb=9\ \text{cm}b=9 cm and c=12 cmc=12\ \text{cm}c=12 cm are given, with included angle A^=52∘\widehat{A}=52^\circA=52∘. (a) Find the side aaa. (b) Find the area of the triangle. (c) Find the angle B^\widehat{B}B. Solution (a) Cosine rule: a2=b2+c2−2bccosA^=81+144−216cos52∘=225−132.98=92.02a^2=b^2+c^2-2bc\cos\widehat{A}=81+144-216\cos 52^\circ=225-132{.}98=92{.}02a2=b2+c2−2bccosA=81+144−216cos52∘=225−132.98=92.02, so a≈9.59a\approx 9{.}59a≈9.59 cm. (b) S=12bcsinA^=12⋅9⋅12⋅sin52∘≈54⋅0.7880≈42.55 cm2S=\dfrac12 bc\sin\widehat{A}=\dfrac12\cdot 9\cdot 12\cdot\sin 52^\circ\approx 54\cdot 0{.}7880\approx 42{.}55\ \text{cm}^2S=21bcsinA=21⋅9⋅12⋅sin52∘≈54⋅0.7880≈42.55 cm2. (c) Sine rule: sinB^b=sinA^a⇒sinB^=9sin52∘9.59≈0.7393⇒B^≈47.66∘\dfrac{\sin\widehat{B}}{b}=\dfrac{\sin\widehat{A}}{a}\Rightarrow \sin\widehat{B}=\dfrac{9\sin 52^\circ}{9{.}59}\approx 0{.}7393\Rightarrow \widehat{B}\approx 47{.}66^\circbsinB=asinA⇒sinB=9.599sin52∘≈0.7393⇒B≈47.66∘. a≈9.59 cm; S≈42.55 cm2; B^≈47.66∘\boxed{a\approx 9{.}59\ \text{cm};\ S\approx 42{.}55\ \text{cm}^2;\ \widehat{B}\approx 47{.}66^\circ}a≈9.59 cm; S≈42.55 cm2; B≈47.66∘