Answer:
\[\angle A\text{ +}\angle B\text{ + }\angle C=180{}^\circ \] 1: 2: 1 = \[180{}^\circ \] Therefore, 1+2+1 = 4 \[\angle ABC=\frac{1}{4}\times 180{}^\circ \] \[=\frac{180{}^\circ }{4}={{45}^{\circ }}\] \[\angle BCA=\frac{5}{4}\times {{180}^{\circ }}\] \[=\frac{180{}^\circ \times 2}{4}\] \[=\text{ }45{}^\circ \times 2=90{}^\circ \] \[\angle CAB=1\times 180{}^\circ =\frac{180{}^\circ }{4}=45{}^\circ \]1 Classification: (i) Isosceles triangle (ii) Right angled triangle
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