Find the value of unknown x in the following diagrams: |
(a) |
(b) |
(c) |
(d) |
Answer:
(a) The sum of interior angles of a triangle is\[180{}^\circ \] \[\therefore \] \[\angle A+\angle B+\angle C=180{}^\circ \] \[x+50{}^\circ +60{}^\circ =180{}^\circ \] \[x+110{}^\circ =180{}^\circ \] \[x={{180}^{\circ }}-{{110}^{\circ }}={{70}^{\circ }}\] (b) The sum of interior angles of a triangle is\[180{}^\circ \] \[\therefore \] \[\angle P+\angle Q+\angle R=180{}^\circ \] \[90{}^\circ +30{}^\circ +x{}^\circ =180{}^\circ \] \[120{}^\circ +x{}^\circ =180{}^\circ \] \[{{x}^{\circ }}={{180}^{\circ }}-{{120}^{\circ }}\] \[x{}^\circ =60{}^\circ \] (c) The sum of interior angles of a triangle is\[180{}^\circ \] \[\angle X+\angle Y\text{+}\angle Z=180{}^\circ \] \[30{}^\circ +110{}^\circ +x{}^\circ =180{}^\circ \] \[140{}^\circ +x{}^\circ =180{}^\circ \] \[x{}^\circ =180{}^\circ -140{}^\circ \] \[x{}^\circ =40{}^\circ \] (d) Since, the sum of all 3 interior angles of a triangle is \[180{}^\circ \]. \[\therefore \] \[50{}^\circ +x{}^\circ +x{}^\circ =180{}^\circ \] \[50{}^\circ +2x{}^\circ =180{}^\circ \] \[2x{}^\circ =180{}^\circ -50{}^\circ \] \[2x{}^\circ =130{}^\circ \] \[x{}^\circ =\frac{130{}^\circ }{2}\] \[x{}^\circ =65{}^\circ \]
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