0 mehakNewbie Asked: June 24, 20232023-06-24T20:29:19+05:30 2023-06-24T20:29:19+05:30In: CBSE 3. In fig. 5.37, find tan P and cot R. Is tan P = cot R? 0 Help with this question. Which formula used? Class 10th cbse rd sharma trigonometric identites rd sharma class 10thtrigonometric identities Share Facebook 1 Answer Voted Oldest Recent mehak Newbie 2023-06-25T19:48:20+05:30Added an answer on June 25, 2023 at 7:48 pm Solution: By using Pythagoras theorem in △PQR, we have PR2 = PQ2 + QR2 Putting the length of given side PR and PQ in the above equation, 132 = 122 + QR2 QR2 = 132 – 122 QR2 = 169 – 144 QR2 = 25 QR = √25 = 5 By definition, tan P = Perpendicular side opposite to P/ Base side adjacent to angle P tan P = QR/PQ tan P = 5/12 ………. (1) And, cot R= Base/Perpendicular cot R= QR/PQ cot R= 5/12 …. (2) When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal. Therefore, L.H.S of both equations should also be equal. ∴ tan P = cot R Yes, tan P = cot R = 5/12 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions 16. A copper sphere of radius 3 cm is melted and recast into a right circular cone of ... 17. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of ... 18. The diameters of the internal and external surfaces of a hollow spherical shell are 10cm and 6 ... 19. How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a ... 20. The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into ...
Solution:
By using Pythagoras theorem in △PQR, we have
PR2 = PQ2 + QR2
Putting the length of given side PR and PQ in the above equation,
132 = 122 + QR2
QR2 = 132 – 122
QR2 = 169 – 144
QR2 = 25
QR = √25 = 5
By definition,
tan P = Perpendicular side opposite to P/ Base side adjacent to angle P
tan P = QR/PQ
tan P = 5/12 ………. (1)
And,
cot R= Base/Perpendicular
cot R= QR/PQ
cot R= 5/12 …. (2)
When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.
Therefore, L.H.S of both equations should also be equal.
∴ tan P = cot R
Yes, tan P = cot R = 5/12