Question

5. Use trigonometry to find all sides and angles in the triangles below 10 in 6 m 35 3 m

Answer 1

10 in 6 m 35 3 m

a) In triangle ABC, Given that $AC=10\,in$ , $\angle B=90\degree$ , $\angle C=35\degree$

$\sin 35\degree=\frac{AB}{AC}\Rightarrow AB=AC\sin35\degree$ , that is $AB=10\sin35\degree=5.74\,in$

$\cos35\degree=\frac{BC}{AC}\Rightarrow BC=AC\cos35\degree$ , that is $BC=10\cos35\degree=8.19\,in$

$\angle A+\angle B+\angle C=180\degree$

$\angle A+90\degree+35\degree=180\degree$

$\Rightarrow \angle A=55\degree$

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b) In triangle PQR , $QR=3\,m$ , $PR=6\,m$ , $\angle R=90\degree$

From Pythagoras theorem , $PQ^2=QR^2+PR^2$

$PQ^2=3^2+6^2\Rightarrow PQ=6.71\,m$

$\sin\angle Q=\frac{PR}{PQ}\Rightarrow \angle Q=\sin^{-1}\left ( \frac{PR}{PQ} \right )=\sin^{-1}\left ( \frac{6}{6.71} \right )=63.4\degree$

$\angle P+\angle Q+\angle R=180\degree$

$\angle P+63.4\degree+90\degree=180\degree$

$\Rightarrow \angle P=26.6\degree$

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