{"id":372,"date":"2022-04-13T10:48:41","date_gmt":"2022-04-13T10:48:41","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=372"},"modified":"2024-11-28T10:37:16","modified_gmt":"2024-11-28T10:37:16","slug":"s-9-solve-a-right-triangle","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/s-9-solve-a-right-triangle\/","title":{"rendered":"S.9 Solve a right triangle"},"content":{"rendered":"\n

Solve a right triangle<\/strong><\/h2>\n\n\n\n

key notes :<\/strong><\/p>\n\n\n\n

The sine (sin) of an angle in a right triangle is a ratio. It is the length of the opposite leg (opp) divided by the length of the hypotenuse (hyp). The opposite leg is the side across from the specified angle and the hypotenuse is the side across from the right angle.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Learn with an example<\/strong><\/p>\n\n\n\n

\n
\n

Solve the right triangle.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Write your answers in simplified, rationalised form. Do not round.<\/strong><\/p>\n\n\n\n

BC =
CD =
\u2220B =______\u00b0<\/p>\n<\/div><\/div>\n\n\n\n

Find sin45\u00b0.<\/strong><\/p>\n\n\n\n

To calculate sin45\u00b0, first draw the 45\u00b0-45\u00b0-90\u00b0 special right triangle.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Calculate sin 45\u00b0.<\/p>\n\n\n\n

sin 45\u00b0 = opp\/hyp Definition of sine<\/p>\n\n\n\n

= 1\/\u221a2 Plug in opp=1 and hyp= \u221a2<\/p>\n\n\n\n

= 1\/\u221a2 . \u221a2\/\u221a2 Rationalise the denominator<\/p>\n\n\n\n

= \u221a2\/2 Multiply<\/p>\n\n\n\n

So, sin4 5\u00b0= \u221a2\/2 .<\/p>\n\n\n\n

Find BC.<\/strong><\/p>\n\n\n\n

Use the definition of sine (sin) to write an equation, then solve for BC.<\/p>\n\n\n\n

sin 45\u00b0 = BC \/ 4\u221a2 Given<\/p>\n\n\n\n

\u221a2\/2 = BC \/ 4\u221a2 Substitute sin 45\u00b0= \u221a2\/2<\/p>\n\n\n\n

4\u221a2 . \u221a2\/2 = BC Multiply both sides by 4\u221a2<\/p>\n\n\n\n

4 = BC Simplify<\/p>\n\n\n\n

So, BC=4.<\/strong><\/p>\n\n\n\n

Find CD.<\/strong><\/p>\n\n\n\n

Use Pythagoras’ theorem to find CD.<\/p>\n\n\n\n

CD2<\/sup>+BC2<\/sup>= BD2<\/sup> Pythagoras’ Theorem<\/p>\n\n\n\n

CD2<\/sup>+42<\/sup>= (4\u221a2)2<\/sup> Plug in BC=4 and BD=4\u221a2<\/p>\n\n\n\n

CD2<\/sup>+16= 32 Square<\/p>\n\n\n\n

CD2<\/sup>= 16 Subtract 16 from both sides<\/p>\n\n\n\n

CD= 4 Take the square root of both sides<\/p>\n\n\n\n

So, CD=4.<\/strong><\/p>\n\n\n\n

Find \u2220B.<\/strong><\/p>\n\n\n\n

\u2220B and \u2220D are the acute angles of a right triangle, so they are complementary. Write an equation setting the sum of their measures equal to 90\u00b0, and solve for \u2220B.<\/p>\n\n\n\n

\u2220B+\u2220D= 90\u00b0<\/p>\n\n\n\n

\u2220B+45\u00b0 = 90\u00b0 Plug in \u2220D=45\u00b0<\/p>\n\n\n\n

\u2220B = 45\u00b0 Subtract 45\u00b0 from both sides<\/p>\n\n\n\n

So, \u2220B=45\u00b0.<\/p>\n\n\n\n

To summarize, BC=4, CD=4 <\/strong>and \u2220B=45\u00b0<\/strong>.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

Solve the right triangle.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Write your answers in simplified, rationalised form. Do not round.<\/strong><\/p>\n\n\n\n

GH =
FH =
\u2220G =______\u00b0<\/p>\n<\/div><\/div>\n\n\n\n

Find sin60\u00b0.<\/strong><\/p>\n\n\n\n

To calculate sin60\u00b0, first draw the 30\u00b0-60\u00b0-90\u00b0 special right triangle.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Calculate sin60\u00b0.<\/p>\n\n\n\n

sin60\u00b0 = opp\/hyp Definition of sine<\/p>\n\n\n\n

= \u221a3\/2 Plug in opp=\u221a3 and hyp=2<\/p>\n\n\n\n

So, sin60\u00b0=\u221a3\/2 .<\/p>\n\n\n\n

Find GH.<\/strong><\/p>\n\n\n\n

Use the definition of sine (sin) to write an equation, then solve for GH.<\/p>\n\n\n\n

sin60\u00b0 = GH\/6 Given<\/p>\n\n\n\n

\u221a3\/2 = GH\/6 Substitute sin60\u00b0=\u221a3\/2<\/p>\n\n\n\n

6 . \u221a3\/2 = GH Multiply both sides by 6<\/p>\n\n\n\n

3\u221a3= GH Simplify<\/p>\n\n\n\n

So, GH = 3\u221a3<\/strong>.<\/p>\n\n\n\n

Find FH.<\/strong><\/p>\n\n\n\n

Use Pythagoras’ theorem to find FH.<\/p>\n\n\n\n

FH2<\/sup>+GH2<\/sup> = FG2 <\/sup> Pythagoras’ Theorem<\/p>\n\n\n\n

FH2<\/sup>+(3\u221a32<\/sup>)= 62<\/sup> Plug in GH=3\u221a3 and FG=6<\/p>\n\n\n\n

FH2<\/sup>+27= 36 Square<\/p>\n\n\n\n

FH2<\/sup> = 9 Subtract 27 from both sides<\/p>\n\n\n\n

FH= 3 Take the square root of both sides<\/p>\n\n\n\n

So, FH=3<\/strong>.<\/p>\n\n\n\n

Find \u2220G.<\/strong><\/p>\n\n\n\n

\u2220F and \u2220G are the acute angles of a right triangle, so they are complementary. Write an equation setting the sum of their measures equal to 90\u00b0, and solve for \u2220G.<\/p>\n\n\n\n

\u2220F+\u2220G= 90\u00b0<\/p>\n\n\n\n

60\u00b0+\u2220G= 90\u00b0 Plug in \u2220F=60\u00b0<\/p>\n\n\n\n

\u2220G= 30\u00b0 Subtract 60\u00b0 from both sides<\/p>\n\n\n\n

So, \u2220G=30\u00b0.<\/p>\n\n\n\n

To summarize, GH=3\u221a3, FH=3 and \u2220G=30\u00b0.<\/p>\n<\/div><\/div>\n\n\n\n

let’s practice!<\/strong><\/p>\n\n\n\n

\n
\n
\"\"<\/a><\/figure>\n<\/div>\n\n\n\n
\n
\"\"<\/a><\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Solve a right triangle key notes : The sine (sin) of an angle in a right triangle is a ratio. It is the length of the opposite leg (opp) divided by the length of the hypotenuse (hyp). The opposite leg is the side across from the specified angle and the hypotenuse is the side acrossContinue reading “S.9 Solve a right triangle”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-372","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/372","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=372"}],"version-history":[{"count":24,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/372\/revisions"}],"predecessor-version":[{"id":14954,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/372\/revisions\/14954"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=372"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}