I want to know solution
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Answered by
11
HELLO DEAR,
GIVEN THAT:-
PR = 13cm
PQ = 12cm
QR = ?
IN ∆ PQR , < Q = 90°
use Pythagoras theorem,
PR² = PQ² + QR²
(13)² = (12)² + QR²
QR² = 169 - 144
QR² = 25
QR = 5
now,
tanp = QR / PQ = 5/12
AND,
cotR = QR / PQ = 5/12
SO ,
tanp - cotR = 5/12 - 5/12
=> 0
I HOPE ITS HELP YOU DEAR,
THANKS
GIVEN THAT:-
PR = 13cm
PQ = 12cm
QR = ?
IN ∆ PQR , < Q = 90°
use Pythagoras theorem,
PR² = PQ² + QR²
(13)² = (12)² + QR²
QR² = 169 - 144
QR² = 25
QR = 5
now,
tanp = QR / PQ = 5/12
AND,
cotR = QR / PQ = 5/12
SO ,
tanp - cotR = 5/12 - 5/12
=> 0
I HOPE ITS HELP YOU DEAR,
THANKS
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Answered by
15
Hey mate,
We know that tan x = perpendicular / base
cot x = base / perpendicular
We Could have found the value Hypotenuse
And substituted the values but before that we got our solution
i.e = 0
See the images for clear understanding
Hope u understand!!
We know that tan x = perpendicular / base
cot x = base / perpendicular
We Could have found the value Hypotenuse
And substituted the values but before that we got our solution
i.e = 0
See the images for clear understanding
Hope u understand!!
Attachments:
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