Solve ques 9 in the attachment
not getting ans: 9 / 2 :-(
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Hey!!!!!
Good Evening
Difficulty Level : Average
Chances of being asked in Board : 70%
____________________
So for the figure refer to the attachment
Let's solve
First of all by Pythagoras theorem we will find
QR = √225 - 100
=> QR = √125
=> QR = 5√5
Now tanP = QR/PQ (opposite/base)
=> tanP = 5√5/10
=> tanP = √5/2
Similarly secP = PR/PQ (hypotenuse/base)
=> secP = 3/2
Using secP and tanP in the To Find
= tan²P + sec²P + 1
= 5/4 + 9/4 + 1
= (5 + 9 + 4)/4
= 18/4
= 9/2 <<<<<<<< Answer
___________________
Hope this helps ✌️
Good Night
Jai Hind
Good Evening
Difficulty Level : Average
Chances of being asked in Board : 70%
____________________
So for the figure refer to the attachment
Let's solve
First of all by Pythagoras theorem we will find
QR = √225 - 100
=> QR = √125
=> QR = 5√5
Now tanP = QR/PQ (opposite/base)
=> tanP = 5√5/10
=> tanP = √5/2
Similarly secP = PR/PQ (hypotenuse/base)
=> secP = 3/2
Using secP and tanP in the To Find
= tan²P + sec²P + 1
= 5/4 + 9/4 + 1
= (5 + 9 + 4)/4
= 18/4
= 9/2 <<<<<<<< Answer
___________________
Hope this helps ✌️
Good Night
Jai Hind
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