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In `DeltaABC`, with usual notations, if `cosA=(sinB)/(sinC)`, then the triangle isA. Acute angled triangleB. Equilateral triangleC. Obtuse angled triangleD. Right angled triangle |
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Answer» Correct Answer - D We have, cos A=`(sinB)/(sinC)` `rArr(b^(2)+c^(2)-a^(2))/(2bc)=b/c(because(sinA)/a=(sinB)/b=(sinC)/c=k)` `rArrb^(2)+c^(2)-a^(2)=2b^(2)` `rArrc^(2)-a^(2)=b^(2)` `rArrc^(2)=a^(2)+b^(2)` `rArrDelta` ABC right angled triangle at `angle`C. |
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