Trigonometry - Study Mode

[#311] In a triangle ABC, ∠ABC = 75° and ∠ACB = $$frac{{{pi ^c}}}{4},$$ xa0the circular measure of ∠BAC is?
Correct Answer

(B) $$frac{pi }{3}$$ radian

Explanation

Solution: $$eqalign{
& frac{{{pi ^c}}}{4}{ ext{ = }}frac{{{{180}^ circ }}}{4}{ ext{ = }}{45^ circ } cr
& angle { ext{BAC}} = {180^ circ } - {75^ circ } - {45^ circ } = {60^ circ } cr
& {180^ circ } o pi cr
& {1^ circ } o frac{pi }{{{{180}^ circ }}} cr
& {60^ circ } o frac{pi }{{{{180}^ circ }}} imes {60^ circ } = frac{pi }{3}{ ext{ radian}} cr} $$

[#312] If $$ heta $$ be acute angle and $$cos heta = frac{{15}}{{17}}{ ext{,}}$$ xa0 then the value of $${ ext{cot}}left( {{{90}^ circ } - heta }
ight)$$ xa0 is?
Correct Answer

(B) $$frac{8}{{15}}$$

Explanation

Solution: $$cos heta = frac{{15 o { ext{Base}}}}{{17 o { ext{Hypo}}}}$$ $$eqalign{
& { ext{Perpendicular = 8}} cr
& Rightarrow { ext{cot}}left( {{{90}^ circ } - heta }
ight) cr
& Rightarrow { ext{tan}} heta = frac{8}{{15}}left[ { herefore an heta = frac{{ ext{P}}}{{ ext{B}}}}
ight] cr} $$

[#313] If a right-angled triangle XYZ right-angled at Y. If XY = $${ ext{2}}sqrt 6 $$ xa0and XZ - YZ = 2, then secX + tanX is?
Correct Answer

(B) $$sqrt 6 $$

Explanation

Solution: $$eqalign{
& XZ - YZ = 2 cr
& h - P = 2 ,........{ ext{(i)}} cr
& {h^2} = {(2sqrt 6 )^2} + {P^2} cr
& {h^2} - {P^2} = {left( {2sqrt 6 }
ight)^2} cr
& (h - P)(h + P) = 4 imes 6 cr
& (2)(h + P) = 24 cr
& h + P = 12 ,..........{ ext{(ii)}} cr
& { ext{Adding eq }}left( { ext{i}}
ight){ ext{ and }}left( {{ ext{ii}}}
ight) cr
& 2h = 14 cr
& h = 7 ,{ ext{and }}P = 5 cr
& herefore secX + tanX cr
& = frac{h}{{XY}} + frac{P}{{XY}} cr
& = frac{7}{{2sqrt 6 }} + frac{5}{{2sqrt 6 }} cr
& = frac{{12}}{{2sqrt 6 }} cr
& = frac{6}{{sqrt 6 }} cr
& = sqrt 6 cr} $$

[#314] In ΔABC, ∠B = 90° and AB : BC = 2 : 1, then value of (sinA + cotC) = ?
Correct Answer

(B) $$frac{{2 + sqrt 5 }}{{2sqrt 5 }}$$

Explanation

Solution: $$eqalign{
& { ext{AC}} = sqrt {{2^2} + {1^2}} = sqrt 5 cr
& { ext{sin A}} + operatorname{cotC} cr
& frac{{{ ext{BC}}}}{{{ ext{AC}}}} + frac{{{ ext{BC}}}}{{{ ext{AB}}}} cr
& frac{1}{{sqrt 5 }} + frac{1}{2} cr
& Rightarrow frac{{2 + sqrt 5 }}{{2sqrt 5 }} cr} $$

[#315] If sin21° = $$frac{x}{y}{ ext{,}}$$xa0 then sec21° - sin69° is equal to?
Correct Answer

(A) $$frac{{{x^2}}}{{ysqrt {{y^2} - {x^2}} }}$$

Explanation

Solution: $$eqalign{
& { ext{In }}vartriangle { ext{ABC sin2}}{{ ext{1}}^ circ }{ ext{ = }}frac{x}{y} cr
& { ext{AB}} = x cr
& { ext{AC}} = y cr
& { ext{BC}} = sqrt {{y^2} - {x^2}} cr
& Rightarrow { ext{sec2}}{{ ext{1}}^ circ } - sin {69^ circ } cr
& Rightarrow frac{{{ ext{AC}}}}{{{ ext{BC}}}} - frac{{{ ext{BC}}}}{{{ ext{AC}}}} cr
& Rightarrow frac{{{{left( {{ ext{AC}}}
ight)}^2} - {{left( {{ ext{BC}}}
ight)}^2}}}{{left( {{ ext{BC}}}
ight)left( {{ ext{AC}}}
ight)}} = frac{{{y^2} - {{left( {sqrt {left( {{y^2} - {x^2}}
ight)} }
ight)}^2}}}{{ysqrt {{y^2} - {x^2}} }} cr
& Rightarrow frac{{{y^2} - {y^2} + {x^2}}}{{ysqrt {{y^2} + {x^2}} }} = frac{{{x^2}}}{{ysqrt {{y^2} - {x^2}} }} cr} $$