Line SA is the perpendicular bisector of Line TR. Measure angle TSR = 120. Which other statement is true? Line TR is the perpendicular bisector of Line SA. Angle STA is congruent to Angle TAR. Line SA is the angle bisector of Angle TAR. Line TR is congruent to Line SA.

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all 3 angles of a triangle is 180 degrees to get the third angle to be 30 degrees. hope it helped :3


Correct answer is C. If this is all you need, you could ignore the rest of solution :) As i said measure ∡ TSR ≠ 120, but we could ignore it. Let O - intersection of TR and SA. A. False. |SO| = 6 and |OA| = 12 so |SO| ≠ |OA| and TR is not the perpendicular bisector of SA. D. False. Two line segments are congruent if they have the same length, but: |TR| = 8 + 8 = 16 |SA| = 6 + 12 = 18 |TR| ≠ |SA| so they are not congruent. C. True TR and SA are perpendicular, so triangles TOA and ROA are right triangles. Moreover they are congruent (same side |AO| = 12, same angle = 90 and same side |TO| = |OR| = 8). But this means that angles TAS and RAS are congruent and have the same measure. So SA is angle bisector of ∡TAR. B. False Here we could use angle TSR, but we can't, so it wil be little harder. First calculate areas of triangles STA and TAR. [latex]A_{STA}=\dfrac{1}{2}(6+12)\cdot8=\dfrac{18\cdot8}{2}=72\\\\\\A_{TAR}=\dfrac{1}{2}(8+8)\cdot12=\dfrac{16\cdot12}{2}=96[/latex] On the other hand, we know that: [latex]|TA|=|AR|=a\qquad |ST|=b\\\\\\A_{STA}=\dfrac{1}{2}|ST|\cdot|TA|\cdot\sin(\angle STA)=\dfrac{1}{2}ab\sin(\angle STA)=72\\\\\\\dfrac{1}{2}ab\sin(\angle STA)=72\quad|\cdot2\\\\\\\boxed{ab\sin(\angle STA)=144}\\\\\\\\ A_{TAR}=\dfrac{1}{2}|TA|\cdot|AR|\cdot\sin(\angle TAR)=\dfrac{1}{2}a^2\sin(\angle TAR)=96\\\\\\\dfrac{1}{2}a^2\sin(\angle TAR)=96\quad|\cdot2\\\\\\\boxed{a^2\sin(\angle TAR)=192}[/latex] Now assume that angle STA is congruent to angle TAR. Then of course m∡STA = m∡ TAR and [latex]\sin(\angle STA)=\sin(\angle TAR)[/latex]. We have: [latex]\dfrac{ab\sin(\angle STA)}{a^2\sin(\angle TAR)}=\dfrac{144}{192}\quad\qquad\sin(\angle STA)=\sin(\angle TAR)\\\\\\\dfrac{ab}{a^2}=\dfrac{3}{4}\\\\\\ \boxed{\dfrac{b}{a}=\dfrac{3}{4}}[/latex] But from Pythagorean theorem we know that: [latex]b^2=6^2+8^2=36+64=100\\\\b=\sqrt{100}=10\\\\\\a^2=8^2+12^2=64+144=208\\\\a=\sqrt{208}=4\sqrt{13}[/latex] and: [latex]\dfrac{b}{a}=\dfrac{10}{4\sqrt{13}}=\dfrac{5}{2\sqrt{13}}\neq\dfrac{3}{4}=\dfrac{b}{a}[/latex] We got contradiction (two different values for b/a), so our assumption that angle STA is congruent to angle TAR was incorrect and B is false. And why m∡TSR ≠ 120? If m∡TSR will be 120, then m∡TSO = 60. We know from definition of tan(x), that: [latex]\tan(\angle TSO)=\dfrac{8}{6}=\dfrac{4}{3}[/latex] but [latex]\tan(\angle TSO)=\tan(60)=\sqrt{3}[/latex] So m∡TSO ≠ 60 and m∡TSR = 2*m∡TSO ≠ 120

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