In Delta ABC,AB = AC and angle B = 75^{circ}, | vedclass

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Question

(C) In $\Delta ABC$,it is given that $AB = AC$.According to the Isosceles Triangle Theorem,the angles opposite to equal sides of a triangle are equal.

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) In $\Delta ABC$,it is given that $AB = AC$.According to the Isosceles Triangle Theorem,the angles opposite to equal sides of a triangle are equal.Therefore,the angle opposite to side $AB$ is $\angle C$ and the angle opposite to side $AC$ is $\angle B$.Since $AB = AC$,it follows that $\angle C = \angle B$.Given that $\angle B = 75^{\circ}$,we have $\angle C = 75^{\circ}$.

In $\Delta ABC$,$AB = AC$ and $\angle B = 75^{\circ}$,then $\angle C = \dots$ (in $^{\circ}$)

Similar Questions

In $\Delta ABC$,$AD$,$BE$,and $CF$ are its medians. Prove that $AB + BC + CA > AD + BE + CF$.

In $\Delta ABC$,$AB = 8 \text{ cm}$ and $BC = 10 \text{ cm}$. Then the perimeter of $\Delta ABC$ is less than how many centimeters?

In $\Delta PQR$,$\angle P : \angle Q : \angle R = 5 : 2 : 2$. Find the measure of each angle of $\Delta PQR$.

$S$ is any point on side $QR$ of a $\triangle PQR$. Show that: $PQ + QR + RP > 2 \, PS$.

In $\Delta PQR$,$\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$,then find $\angle R$. (in $^{\circ}$)

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