Question

In triangle PQR, if PQ:QR:PR = 8:15:17, then evaluate
1) cos P . cos R - sin P . sin R

2) tan P - tan R / 1+tan P . tan R
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Answer 1

Solution :

1) CosP CosR - SinP SinR

=> [ ( 8/17 ) × (15/17) ] - [ (15/17) × (8/17) ]

=> 120/289 - 120/289

=> 0

2) [(15/8)-(15/8)]/[1+(8/15)(15/8)]

=> [(225-64)/120]/2

=> 161/120 × 2

=> 161/60

For more, refer to the attachment.

Extra :

• SinA = P/H
• CosA = B/H
• TanA = P/B
• CosecA = H/P
• sec A = H/B
• CotA = B/P

Answer 2

$\huge\star{\green{\underline{\mathfrak{AnsweR:-}}}}$

Given:-

PQ:QR:PR = 8:15:17

To Find:-

• cos P . cos R - sin P . sin R

• tan P - tan R / 1+tan P . tan R

$\rule{200}2$

Solution:-

Given, PQ:QR:PR = 8:15:17

So, PQ=8k, QR=15k, PR=17k.

So,

$\hookrightarrow\tt{cos\:P=\dfrac{8}{17}}$

$\hookrightarrow\tt{cos\:R=\dfrac{15}{17}}$

$\hookrightarrow\tt{sin\:P=\dfrac{15}{17}}$

$\hookrightarrow\tt{sin\:R=\dfrac{8}{17}}$

$\rule{200}2$

So answer for first question:-

$\mapsto\tt{cos P \times cos R - sin P \times sin R}$

$\mapsto\tt{\dfrac{8}{17}\times\dfrac{15}{17}-\dfrac{15}{17}\times \dfrac{8}{17}}$

$\mapsto\tt{\dfrac{120}{289}=\dfrac{120}{289}}$

$\mapsto\tt{\boxed{\red{O}}}$

$\rule{200}2$

Now,

$\hookrightarrow\tt{tan\:P=\dfrac{15}{8}}$

$\hookrightarrow\tt{tan\:R=\dfrac{8}{15}}$

$\rule{200}2$

So, answer for the answer of 2nd question:-

$\mapsto\tt{\dfrac{tan\:P-tan\:R}{1+tan\:P\times tan\:R}}$

$\mapsto\tt{\dfrac{\dfrac{15}{8}-\dfrac{8}{15}}{1+\bigg(\dfrac{\cancel{15}}{\cancel{8}}\times \dfrac{\cancel{8}}{\cancel{15}}\bigg)}}$

$\mapsto\tt{\dfrac{\dfrac{15}{8}-\dfrac{8}{15}}{1+(1)}}$

$\mapsto\tt{\dfrac{\dfrac{225-64}{120}}{2}}$

$\mapsto\tt{\dfrac{161}{\cancel{120}\:\:60}\times\dfrac{\cancel{2}}{1}}$

$\mapsto\tt{\boxed{\red{\dfrac{161}{60}}}}$

$\rule{200}2$

$\rule{200}2$

Some Formulas:-

• $\tt{sin\:\theta=\dfrac{P}{H}}$

• $\tt{cos\:\theta=\dfrac{B}{H}}$

• $\tt{tan\:\theta=\dfrac{P}{B}}$

• $\tt{cosec\:\theta=\dfrac{H}{P}}$

• $\tt{sec\:\theta=\dfrac{H}{B}}$

• $\tt{cot\:\theta=\dfrac{B}{P}}$

$\rule{200}4$