Question

URGENT - PLEASE HELP !!!

Find the measure of angle TRS and angle RST

BTW it says that it’s a square but i’m not sure how bc it looks like a rhombus

Answer 1

`\sf\\\angle TRS=58^o\ and\ \angle RST=64^o`

Step-by-step explanation:

`\sf\\\textsf{I am considering this figure to be a rhombus. Because if it was a square, then }\\(6x+4)^o\textsf{ and }(5x-13)^o\textsf{ will be both equal to }45^o\textsf{ so therefore there will be two}\\\textsf{values of x which is not possible.}`

`\textsf{The solution to this problem is as follows:}\\`

`\sf\\1.\ \\(i)\ UR=RS=ST=UT\ \ \ [\textsf{All sides of rhombus are equal.}]\\(ii)\ \triangle RUS\ and\ \triangle URT\ are\ isosceles.\ [UR=RS\ and\ UR=UT]\\\\2.\ \textsf{As base angles of isosceles triangle are equal,}\\(i)\ \angle RUS=\angle RSU=(5x-13)^o\\(ii)\ \angle URT=\angle UTR=(6x+4)^o\\\\3.\ \textsf{Since alternate angles are equal,}\\(i)\ \angle RSU=\angle SUT=(5x-13)^o\\(ii)\ \angle UTR=\angle TRS=(6x+4)^o`

`\sf\\(4)\ \angle R+\angle U=180^o\ \ \ [\textsf{Co-interior angles are supplementary.}]\\or,\ (\angle URT+\angle TRS)+(\angle RUS+\angle SUT)=180^o\\or,\ (6x+4^o+6x+4^o)+(5x-13^o+5x-13^o)=180^o\\or,\ 22x-18^o=180^o\\or,\ 22x=198^o\\or,\ x=9^o`

`\sf\\5.\ \angle TRS=6x+4^o=6(9^o)+4^o=58^o\ \ \ [\textsf{From step 3. (ii)}]\\\\6.\ \angle RUT=\angle RUS+\angle SUT=5x-13^o+5x-13^o=10x-26^o=10(9)-26^o=64^o\\\\7.\ \angle RST=\angle RUT=64^o`