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ABC is an isosceles triangle. The side BA is produced to D such that AB = AD. If \(\angle ABC=35^o \), then \(\angle BCD= \)
\(\triangle PQR \) is an isosceles triangle inscribed in a circle. If \(PQ=PR=12\sqrt5 \ \text{cm},\ QR=24\ \text{cm}, \) then the radius of the circle is
If \(3\angle A=4\angle B=6\angle C \) then \(A:B:C= \)
In \(\triangle ABC \), If \(\angle A\angle B=42^o \) and \(\angle B\angle C=21^o \) then \(\angle B= \)?
In \(\triangle ABC \), side \(BC \) is produced to \(D \). If \(\angle ABC=50^o \) and \(\angle ACD=110^o \) then \(\angle A= \)?
Side BC of \(\triangle \)ABC has been produced to D on left hand side and to E on right hand side such that \(\angle ABD=125^o \) and \(\angle ACE=130^o \) Then \(\angle A= \)?
In
(Exterior angle property)
The sides BC, CA and AB of \(\triangle ABC \) has been produced to D, E, F respectively \(\angle \)BAE+ \(\angle \)CBF+ \(\angle \)ACD = ?
\(EAD\bot BCD \). Ray \(FAC \) cuts ray \(EAD \) at a point. A such that \(\angle EAF=30^o \) In \(\triangle BAC,\angle BAC=x \) and \(\angle ABC=(x+10) \). Then the value of x is
(V.O.A)
Two rays BD and CE intersect at a point A. The side BC of \(\triangle \) ABC has been produced on both sides to points F and G respectively. If \(\angle ABF=x,\ \angle ACG=y \) and \(\angle DAE=z \) then \(z=? \)
(Linear pair)
(Linear pair)
(V.O.A)
If \(\angle OCA=80^o,\angle COA=40^o \) and \(\angle BDO=70^o \) then \( x+y=?\)
(Linear pair)
(V.O.A)
(Linear pair)
In \(\triangle ABC,\ AB=ab,\ AC=\sqrt{a^2+b^2} \) and \(BC=\sqrt{2ab},\ \angle B= \)
Let O be the orthocentre of the triangle ABC. If \( \angle \text{BOC}=140^o\). Then \(\angle BAC= \)
Two medians AD and BE of \(\triangle \) ABC intersect at G at right angle. If AD = 18 cm and BE = 12 cm then the length of BD (in cm) is
In \( \triangle ABC, AD \bot BC\) and \(AD^2=BD.DC \). The measure of \(\angle BAC \) is
Adding equation (1) and (2) we get
In \(\triangle \text{PQR} \), X and Y are two points on PQ and PR respectively such that XY  QR, bisects the \(\triangle \text{ABC} \) in two equal areas. Then the ratio OX : PQ is