Question

$\large\red{\underline{{\boxed{\textbf{question:-}}}}}$
$\large\pink{\underline{{\boxed{\textbf{topic:- \: trigonometry}}}}}$
◆ In ∆OPQ, right angled at P, OP = 7cm and OQ -OP = 1cm. Determine values of all trigonometry ratios in terms of Angle Q.
$\large\green{\underline{{\boxed{\textbf{no \: spams}}}}}$

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Answer 1

$ans$

In triangle OPQ

OQ²= OP²+PQ²

(PQ+1)² = OP²+PQ² [therefore , OQ - PQ = 1 = OQ = 1+PQ ]

PQ²+2PQ + 1= OP²+PQ²

2PQ + 1 = 49

PQ = 24 cm

therefore, OQ - PQ = 1cm

OQ = (PQ + 1 )cm = 25cm

$now \: \sin(q) = \: \frac{op}{oq} \: = \frac{7}{25 } \\ and \: \cos(q) \: = \: \frac{pq}{oq} \: = \frac{24}{25} \: \\ ans \\ hope \: it \: hlps \: uh$

Answer 2

Since the line PQ divides △ABC into two equal parts,

area(△APQ)=area(△BPQC)

⇒  area(△APQ)=area(△ABC)−area(△APQ)

⇒  2area(△APQ)=area(△ABC)

∴   area(△APQ)area(△ABC)=12     ---- ( 1 )

Now, in △ABC and △APQ,

∠BAC=∠PAQ           [ Common angles ]

∠ABC=∠APQ          [ Corresponding angles ]

∴   △ABC∼△APQ           [ By AA similarity ]

∴  area(△APQ)area(△ABC)=AP2AB2

∴  12=AP2AB2           [ From ( 1 ) ]

⇒  APAB=12

⇒  ABAB−BP